Questions Further Statistics B AS

State an assumption about the distribution of the population from which these weights come that is necessary in order to produce this interval.
State the confidence interval which the software gives, in the form $a < \mu < b$.
Explain
- what the label df means,
- how the value of df is calculated for a confidence interval produced using the $t$ distribution.
- State two ways in which the researcher could have obtained a narrower confidence interval.

OCR MEI Further Statistics B AS 2018 June Q2

2 A supermarket sells oranges. Their weights are modelled by the random variable $X$ which has a Normal distribution with mean 345 grams and standard deviation 15 grams. When the oranges have been peeled, their weights in grams, $Y$, are modelled by $Y = 0.7 X$.

Find the probability that a randomly chosen peeled orange weighs less than 250 grams. I randomly choose 5 oranges to buy.
Find the probability that the total weight of the 5 unpeeled oranges is at least 1800 grams.
I peel three of the oranges and leave the remaining two unpeeled. Find the probability that the total weight of the two unpeeled oranges is greater than the total weight of the three peeled ones.

OCR MEI Further Statistics B AS 2018 June Q3

3 The probability density function of the continuous random variable $X$ is given by $$\mathrm { f } ( x ) = \begin{cases} c + x & - c \leqslant x \leqslant 0
c - x & 0 \leqslant x \leqslant c
0 & \text { otherwise } \end{cases}$$ where $c$ is a positive constant.

(A) Sketch the graph of the probability density function.
(B) Show that $c = 1$.
Find $\mathrm { P } \left( X < \frac { 1 } { 4 } \right)$.
Find
- the mean of $X$,
- the standard deviation of $X$.

OCR MEI Further Statistics B AS 2018 June Q4

4 The random variable $X$ has a continuous uniform distribution on [ 0,10 ].

Find $\mathrm { P } ( 3 < X < 6 )$.

Find each of the following.

$\mathrm { E } ( X )$
$\operatorname { Var } ( X )$

Marisa is investigating the sample mean, $Y$, of 8 independent values of $X$. She designs a simulation shown in the spreadsheet in Fig. 4.1. Each of the 25 rows below the heading row consists of 8 values of $X$ together with the value of $Y$. All of the values in the spreadsheet have been rounded to 2 decimal places. \begin{table}[h]

1	A	B	C	D	E	F	G	H	I	J
1	$X _ { 1 }$	$X _ { 2 }$	$X _ { 3 }$	$X _ { 4 }$	$X _ { 5 }$	$X _ { 6 }$	$X _ { 7 }$	$X _ { 8 }$	$Y$
2	6.31	2.45	3.27	3.06	4.16	1.53	0.43	7.99	3.65
3	1.70	1.52	7.10	8.93	6.44	2.70	9.96	7.83	5.77
4	9.15	0.52	4.95	6.99	6.52	3.15	0.81	5.35	4.68
5	0.65	2.71	7.92	9.65	0.50	4.87	6.46	2.67	4.43
6	3.09	6.11	3.96	0.09	0.18	4.67	0.67	6.20	3.12
7	7.06	5.84	1.97	3.60	9.36	1.97	4.48	3.47	4.72
8	1.46	1.57	5.45	0.37	3.76	7.56	8.48	9.12	4.72
9	9.42	1.85	4.91	1.61	1.94	8.00	1.77	5.34	4.36
10	2.98	5.32	2.91	4.12	9.16	1.76	9.97	6.88	5.39
11	2.83	3.44	3.28	7.85	1.00	0.93	8.77	4.03	4.01
12	4.51	0.59	5.84	9.87	8.65	3.94	7.18	0.23	5.10
13	4.49	0.69	3.65	8.78	4.96	8.96	3.77	1.43	4.59
14	6.57	8.08	4.85	6.75	7.92	0.27	9.69	4.04	6.02
15	8.35	1.09	8.63	8.04	7.23	2.12	2.57	9.59	5.95
16	5.24	9.53	6.08	8.21	3.61	7.07	6.65	7.63	6.75
17	7.89	5.50	3.09	0.71	6.47	5.49	6.47	4.95	5.07
18	8.36	7.27	2.35	9.04	0.58	2.26	3.01	7.90	5.10
19	3.76	1.01	9.61	9.65	7.89	9.98	6.28	4.34	6.56
20	9.94	6.84	3.38	5.53	0.26	8.53	5.72	5.12	5.66
21	7.25	9.10	0.34	2.88	4.66	2.65	6.37	7.63	5.11
22	7.18	7.14	5.38	0.04	4.09	6.47	4.96	4.23	4.94
23	8.69	5.04	4.90	2.94	2.00	4.23	4.13	0.97	4.11
24	3.46	6.33	0.48	9.35	0.23	1.18	7.97	6.37	4.42
25	2.37	7.26	7.16	1.24	5.26	2.80	3.55	3.84	4.19
26	2.16	8.30	7.17	3.32	2.96	1.30	9.11	0.31	4.33
27

\captionsetup{labelformat=empty} \caption{Fig. 4.1}

\end{table}

Use the spreadsheet to estimate $\mathrm { P } ( 3 < Y < 6 )$.
Explain why it is not surprising that this estimated probability is substantially greater than the value which you calculated in part (i). Marisa wonders whether, even though the sample size is only 8, use of the Central Limit Theorem will provide a good approximation to $\mathrm { P } ( 3 < Y < 6 )$.
Calculate an estimate of $\mathrm { P } ( 3 < Y < 6 )$ using the Central Limit Theorem. A Normal probability plot of the 25 simulated values of $Y$ is shown in Fig. 4.2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0c58d4d7-10e9-473a-888a-b407ec90bf08-5_800_1291_306_386} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
\end{figure}
Explain what the Normal probability plot suggests about the use of the Central Limit Theorem to approximate $\mathrm { P } ( 3 < Y < 6 )$. Marisa now decides to use a spreadsheet with 1000 rows below the heading row, rather than the 25 which she used in the initial simulation shown in Fig. 4.1. She uses a counter to count the number of values of $Y$ between 3 and 6. This value is 808.
Explain whether the value 808 supports the suggestion that the Central Limit Theorem provides a good approximation to $\mathrm { P } ( 3 < Y < 6 )$. Marisa decides to repeat each of her two simulations many times in order to investigate how variable the probability estimates are in each case.
Explain whether you would expect there to be more, the same or less variability in the probability estimates based on 1000 rows than in the probability estimates based on 25 rows.

OCR MEI Further Statistics B AS 2018 June Q5

5 The flight time between two airports is known to be Normally distributed with mean 3.75 hours and standard deviation 0.21 hours. A new airline starts flying the same route. The flight times for a random sample of 12 flights with the new airline are shown in the spreadsheet (Fig. 5), together with the sample mean. \begin{table}[h]

	A	B	C	D	E	F	G	H	I	J	K	L
1	3.595	3.723	3.584	3.643	3.669	3.697	3.550	3.674	3.924	3.563	3.330	3.706
2
3	Mean	3.638

\captionsetup{labelformat=empty} \caption{Fig. 5}

\end{table} \section*{(i) In this question you must show detailed reasoning.} You should assume that:

the flight times for the new airline are Normally distributed,
the standard deviation of the flight times is still 0.21 hours.

Carry out a test at the $5 \%$ significance level to investigate whether the mean flight time for the new airline is less than 3.75 hours.
(ii) If both of the assumptions in part (i) were false, name an alternative test that you could carry out to investigate average flight times, stating any assumption necessary for this test.
(iii) If instead the flight times were still Normally distributed but the standard deviation was not known to be 0.21 hours, name another test that you could carry out.

OCR MEI Further Statistics B AS 2018 June Q6

6 A company has a large fleet of cars. It is claimed that use of a fuel additive will reduce fuel consumption. In order to test this claim a researcher at the company randomly selects 40 of the cars. The fuel consumption of each of the cars is measured, both with and without the fuel additive. The researcher then calculates the difference $d$ litres per kilometre between the two figures for each car, where $d$ is the fuel consumption without the additive minus the fuel consumption with the additive. The sample mean of $d$ is 0.29 and the sample standard deviation is 1.64 .

Showing your working, find a 95\% confidence interval for the population mean difference.
Explain whether the confidence interval suggests that, on average, the fuel additive does reduce fuel consumption.
Explain why you can construct the interval in part (i) despite not having any information about the distribution of the population of differences.
Explain why the sample used was random.

OCR MEI Further Statistics B AS 2019 June Q1

1 It is known that the red blood cell count of adults in a particular country, measured in suitable units, has mean 4.96 and variance 0.15.

Find the probability that the mean red blood cell count of a random sample of 50 adults from this country is at least 5.00.
Explain how you can find the probability in part (a) despite the fact that you do not know the distribution of red blood cell counts.

OCR MEI Further Statistics B AS 2019 June Q2

2 Leila and Caleb are playing a game, using fair six-sided dice and unbiased coins.

Leila rolls two dice, and her score $L$ is the total of the scores on the two dice.
Caleb spins 4 coins and his score $C$ is three times the number of heads obtained.

The winner of a game is the player with the higher score. If the two scores are equal, the result of the game is a draw. The spreadsheet in Fig. 2 shows a simulation of 20 plays of the game. \begin{table}[h]

	A	B	C	D	E	F	G	H
1	First dice	Second dice	Total (Leila's score) $\boldsymbol { L }$	Coin 1	Coin 2	Coin 3	Coin 4	Caleb's score $\boldsymbol { C }$
2	1	2	3	H	T	T	T	3
3	6	1	7	T	H	T	T	3
4	2	6	8	H	H	T	T	6
5	2	5	7	T	H	H	H	9
6	1	5	6	T	H	T	T	3
7	5	2	7	H	H	H	H	12
8	1	1	2	H	T	H	T	6
9	2	6	8	T	H	T	H	6
10	6	2	8	H	T	H	T	6
11	1	3	4	T	H	H	H	9
12	6	1	7	T	H	T	T	3
13	3	1	4	T	T	T	T	0
14	3	6	9	H	T	H	H	9
15	2	3	5	T	H	H	H	9
16	2	5	7	H	H	H	H	12
17	1	5	6	H	H	T	H	9
18	5	6	11	T	H	H	H	9
19	4	2	6	T	H	H	T	6
20	6	5	11	T	T	H	H	6
21	1	1	2	T	T	T	T	0

\captionsetup{labelformat=empty} \caption{Fig. 2}

\end{table}

Explain why the value of $C$ in row 2 is 3 .
Use the spreadsheet to estimate $\mathrm { P } ( C > 6 )$ and $\mathrm { P } ( L > 6 )$.
Use the spreadsheet to estimate the probability that Leila loses a randomly chosen game.
Explain why your answers to parts (b) and (c) may not be very close to the true values.
Leila claims that the game is fair (that Leila and Caleb each have an equal chance of winning) because both she and Caleb can get a maximum score of 12 and also in the simulation she won exactly $50 \%$ of the games.
Make 2 comments about Leila’s claim.

OCR MEI Further Statistics B AS 2019 June Q3

3 A bus runs from point A on the outskirts of a city, stops at point B outside the rail station, and continues to point C in the city centre.
The journey times for the sections A to B and B to C vary according to traffic conditions, and are modelled by independent Normal distributions with means and standard deviations as shown in the table.

\multirow{2}{*}{}	Journey time (minutes)
\cline { 2 - 3 }	Mean	Standard deviation
A to B	21	3
B to C	29	4

Find the probability that a randomly chosen journey from A to B takes less than the scheduled time of 23 minutes. For every journey, the bus stops for 1 minute when it reaches B to drop off and pick up passengers.
Find the probability that a randomly chosen journey from A to C takes less than the scheduled time of 50 minutes. Mary travels on the bus from the station at B to her workplace at C every working day. You should assume that times for her bus journeys on different days are independent.
Find the probability that the total time taken for her five journeys on the bus in a randomly chosen week is at least $2 \frac { 1 } { 2 }$ hours.
Comment on the assumption that times on different days are independent.

OCR MEI Further Statistics B AS 2019 June Q4

4 The cumulative distribution function of the continuous random variable $X$ is given by
$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 ,
k \left( 12 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 2 ,
1 & x > 2 , \end{cases}$ where $k$ is a constant.

Show that $k = 0.05$.
Find $\mathrm { P } ( 1 \leqslant X \leqslant 1.5 )$.
Find the median of $X$, correct to 3 significant figures.
Find which of the median, mean and mode of $X$ is the largest of the three measures of central tendency.

OCR MEI Further Statistics B AS 2019 June Q5

5 A technician is investigating whether a batch of nylon 66 (a particular type of nylon) is contaminated by another type of nylon.
The average melting point of nylon 66 is $264 ^ { \circ } \mathrm { C }$. However, if the batch is contaminated by the other type of nylon the melting point will be lower. The melting points, in ${ } ^ { \circ } \mathrm { C }$, of a random sample of 8 pieces of nylon from the batch are as follows.
$\begin{array} { l l l l l l l l } 262.7 & 265.0 & 264.1 & 261.7 & 262.9 & 263.5 & 261.3 & 262.6 \end{array}$

Find
- the sample mean,
- the sample standard deviation.
The technician produces a Normal probability plot and carries out a Kolmogorov-Smirnov test for these data as shown in Fig. 5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1db026a2-dffc-4877-b927-247fbf0e7a78-5_560_1358_982_246} \captionsetup{labelformat=empty} \caption{Fig. 5}
\end{figure}
Comment on what the Normal probability plot and the $p$-value of the test suggest about the data.
In this question you must show detailed reasoning. Carry out a suitable test at the $5 \%$ significance level to investigate whether the batch appears to be contaminated with another type of nylon.
Name an alternative test that could have been carried out if the population standard deviation had been known.

OCR MEI Further Statistics B AS 2019 June Q6

6 The label on a pack of strawberries in a large batch states that it holds 250 g of strawberries. A random sample of 40 packs from the batch is selected and software is used to produce a $95 \%$ confidence interval for the mean weight of strawberries per pack. An extract from the software output is shown in Fig. 6. \begin{table}[h]

Sample Mean	248.92
Standard Error	0.61506
Sample Size	40
Confidence Level	0.95
Interval	$248.92 \pm 1.2055$

\captionsetup{labelformat=empty} \caption{Fig. 6}

\end{table}

Explain whether the confidence interval suggests that the mean weight of strawberries per pack in the batch is different from 250 g .
A manager looking at the data says that the conclusion would have been different if a $90 \%$ confidence interval had been used.
Determine whether the manager is correct.
Explain briefly whether or not it is appropriate for the manager to vary the confidence level before coming to any conclusions.
[0pt]
On another occasion, using the same sample size, a 95\% confidence interval for the mean weight of strawberries per pack is [248.05, 249.95].
Find the sample variance in this case.
Explain the meaning of a 95\% confidence interval.

OCR MEI Further Statistics B AS 2022 June Q1

1 Each working day, Beth takes a bus to her place of work. She believes that the mean time that her journey takes is 30 minutes. In order to check this, Beth selects a random sample of 8 journeys. The times in minutes for these 8 journeys are as follows.
$\begin{array} { l l l l l l l l } 31.9 & 28.5 & 35.9 & 31.0 & 30.2 & 34.9 & 28.9 & 31.3 \end{array}$

What assumption does Beth need to make in order to construct a confidence interval for the mean journey time based on the $t$ distribution?
In this question you must show detailed reasoning. Given that the assumption in part (a) is valid, determine a 95\% confidence interval for the mean journey time.
Explain whether the confidence interval suggests that Beth may be correct in the belief that her mean journey time is 30 minutes.

OCR MEI Further Statistics B AS 2022 June Q2

2 The continuous random variable $X$ has cumulative distribution function given by
$F ( x ) = \begin{cases} 0 & x < a ,
\frac { x - a } { b - a } & a \leqslant x \leqslant b ,
1 & x > b , \end{cases}$
where $a$ and $b$ are constants with $0 < \mathrm { a } < \mathrm { b }$.

Find $\mathrm { P } \left( \mathrm { X } < \frac { 1 } { 2 } ( \mathrm { a } + \mathrm { b } ) \right)$.
Sketch the graph of the probability density function of $X$.
Find the variance of $X$ when $a = 2$ and $b = 8$.

OCR MEI Further Statistics B AS 2022 June Q3

3 A local council collects domestic kitchen waste for composting. Householders place their kitchen waste in a 'compost bin' and this is emptied weekly by the council. The average weight of kitchen waste collected per household each week is known to be 3.4 kg . The council runs a campaign to try to increase the amount of kitchen waste per household which is put in the compost bin. After the campaign, a random sample of 40 households is selected and the weights in kg of kitchen waste in their compost bins are measured. A hypothesis test is carried out in order to investigate whether the campaign has been successful, using software to analyse the sample. The output from the software is shown below.
□
Z Test of a Mean
Null Hypothesis $\mu = 3.4$
Alternative Hypothesis $\bigcirc < 0 > 0 \neq$
Sample

Mean	3.565
s	1.05
N	40

Result

Z Test of a Mean

Mean	3.565
S	1.05
SE	0.1660
N	40
Z	0.994
p	0.160

Explain why the test is based on the Normal distribution even though the distribution of the population of amounts of kitchen waste per household is not known.
Using the output from the software, complete the test at the $5 \%$ significance level.
Show how the value of $Z$ in the software output was calculated.
Calculate the least value of the sample mean which would have resulted in the conclusion of the test in part (b) being different. You should assume that the standard error is unchanged.

OCR MEI Further Statistics B AS 2022 June Q4

4 A wood contains a large number of mature beech trees. The diameters in centimetres of a random sample of 10 of these trees are as follows.
$\begin{array} { l l l l l l l l l l } 82.6 & 79.2 & 77.8 & 38.4 & 88.1 & 32.2 & 26.5 & 23.4 & 94.3 & 104.2 \end{array}$ A tree surgeon wants to know if the average diameter of mature beech trees in this wood is 50 cm . The tree surgeon produces a Normal probability plot for these data.
\includegraphics[max width=\textwidth, alt={}, center]{4caa7409-cb32-41da-ad64-012a45753296-4_796_1230_589_230}

Explain why the tree surgeon should not carry out a test based on the $t$ distribution.
Carry out a suitable test at the $5 \%$ significance level to investigate whether the average diameter of mature beech trees in this wood is 50 cm .

OCR MEI Further Statistics B AS 2022 June Q5

5 Layla works at an internet café. Each terminal at the café has its own keyboard, and keyboards need to be replaced whenever faults develop. Layla knows that the number of weeks for which a keyboard lasts before it needs to be replaced can be modelled by the random variable $X$, which has an exponential distribution with mean 20 and variance 400 . She wants to investigate how likely it is that the keyboard at a terminal will need to be replaced at least 3 times within a year (taken as being a period of 52 weeks). Layla designs the simulation shown in the spreadsheet below. Each of the 20 rows below the heading row consists of 3 values of $X$ together with their sum $T$. All of the values in the spreadsheet have been rounded to 1 decimal place.

	A	B	C	D
1	$\mathrm { X } _ { 1 }$	$\mathrm { X } _ { 2 }$	$\mathrm { X } _ { 3 }$	T
2	10.9	21.5	5.3	37.7
3	23.9	52.4	85.3	161.6
4	5.2	10.4	24.0	39.6
5	2.9	14.4	0.8	18.1
6	9.0	43.3	49.7	102.0
7	0.4	16.2	12.4	29.0
8	44.1	39.5	22.1	105.7
9	9.2	43.6	13.9	66.7
10	40.4	10.9	6.1	57.4
11	3.2	54.8	15.7	73.7
12	5.3	6.1	1.6	13.0
13	20.5	28.9	22.9	72.3
14	37.3	2.1	28.6	68.0
15	7.1	13.6	50.1	70.8
16	18.6	2.0	9.3	29.9
17	9.0	1.2	49.9	60.1
18	1.9	9.5	69.8	81.2
19	9.0	2.1	10.4	21.5
20	28.7	1.4	93.8	123.9
21	1.8	2.9	34.8	39.5

Explain why $T$ represents the number of weeks after which the third keyboard at a terminal will need to be replaced.
Use the information in the spreadsheet to write down an estimate of $\mathrm { P } ( T > 52 )$.
Explain how you could obtain a more reliable estimate of $\mathrm { P } ( T > 52 )$.
The internet café has 50 terminals. You are given that faults in keyboards occur independently of each other. Determine an estimate of the probability that the mean number of weeks before which the third keyboard at a terminal needs to be replaced is more than 52 .

OCR MEI Further Statistics B AS 2022 June Q6

6 The length $L$ of a particular type of fence panel is Normally distributed with mean 179.2 cm and standard deviation 0.8 cm . You should assume that the lengths of individual fence panels are independent of each other.

Find the probability that the length of a randomly chosen fence panel is at least 180 cm .
Find the probability that the total length of 5 randomly chosen fence panels is less than 895 cm . The width $W$ of a fence post is Normally distributed with mean 9.8 cm and standard deviation 0.3 cm . A straight fence is constructed using 6 posts and 5 panels with no gaps between them. Fig. 6 shows a view from above of the first two posts, the first panel and the start of the second panel. You should assume that the lengths of fence panels and widths of fence posts are independent. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4caa7409-cb32-41da-ad64-012a45753296-6_213_1522_934_244} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
Determine the probability that the total length of the fence, including the posts, is less than 9.5 m .
State another assumption that is necessary for the calculation of the probability in part (c) to be valid.

OCR MEI Further Statistics B AS 2022 June Q7

7 Many cars have pollen filters to try to remove as much pollen as possible from the passenger compartment. In a test car, the amount of pollen is regularly monitored. The amount of pollen is measured using a scale from 0 to 1 , and is modelled by the continuous random variable $X$ with probability density function given by
$f ( x ) = \begin{cases} k \left( 5 x ^ { 4 } - 16 x ^ { 2 } + 11 x \right) & 0 \leqslant x \leqslant 1 ,
0 & \text { otherwise } , \end{cases}$
where $k$ is a positive constant.

Show that $k = \frac { 6 } { 7 }$.
Determine $\mathrm { P } ( X < \mathrm { E } ( X ) )$.
Verify that the median amount of pollen according to the model lies between 0.417 and 0.418.

OCR MEI Further Statistics B AS 2021 November Q1

1 Over time LED light bulbs gradually lose brightness. For a particular type of LED bulb, it is known that the mean reduction in brightness after 10000 hours is $2.6 \%$. A manufacturer produces a new version of this bulb, which costs less to make, but is claimed to have the same reduction in brightness after 10000 hours as the previous version. In order to check this claim, a random sample of 10 bulbs is selected. For each bulb, the original brightness and the brightness after 10000 hours are measured, in suitable units. A spreadsheet is used to produce a $95 \%$ confidence interval for the mean percentage reduction in brightness. A screenshot of the spreadsheet is shown in Fig. 1. \begin{table}[h]

A		B	C	D	E	F	G	H	1	J	K
1	Original brightness	1075	1121	1106	1095	1101	1109	1114	1123	1108	1115
2	After 10000 hours	1042	1084	1076	1065	1070	1079	1081	1091	1080	1082
3	Percentage reduction	3.07	3.30	2.71	2.74	2.82	2.71	2.96	2.85	2.53	2.96
4
5
6	Sample mean (\%)	2.8650
7	Sample sd (\%)	0.2179
8	SE	0.0689
9	DF	9
10	tvalue	2.262
11	Lower limit	2.709
12	Upper limit	3.021
1.3

\captionsetup{labelformat=empty} \caption{Fig. 1}

\end{table}

State the confidence interval in the form $a < \mu < b$.
Explain whether the confidence interval suggests that the mean percentage reduction in brightness after 10000 hours is different from 2.6\%.
Explain how the value in cell B8 was calculated.
State an assumption necessary for this confidence interval to be calculated.
Explain the advantage of using the same bulbs for both measurements.

OCR MEI Further Statistics B AS 2021 November Q2

2 Natasha is investigating binomial distributions. She constructs the spreadsheet in Fig. 2 which shows the first 3 and last 4 rows of a simulation involving two independent variables, $X$ and $Y$, with distributions $\mathrm { B } ( 10,0.3 )$ and $\mathrm { B } ( 50,0.3 )$ respectively. The spreadsheet also shows the corresponding value of the random variable $Z$, defined by $Z = 5 X - Y$, for each pair of values of $X$ and $Y$. There are 100 simulated values of each of $X , Y$ and $Z$. The spreadsheet also shows whether each value of $Z$ is greater than 6, and cells D103 and D104 show the number of values of $Z$ which are greater than 6 and not greater than 6 respectively. \begin{table}[h]

1	A	B	C	D	E
1	X	Y	$\mathbf { Z } = \mathbf { 5 } \mathbf { X } - \mathbf { Y }$	$\mathbf { Z } > \mathbf { 6 }$
2	4	13	7	Y
3	4	17	3	N
4	3	21	-6	N
5
6
98	1	14	-9	N
99	5	12	13	Y
100	3	18	-3	N
101	3	15	0	N
102
103			Number of Y	19
104			Number of N	81
105

\captionsetup{labelformat=empty} \caption{Fig. 2}

\end{table}

Use the information in the spreadsheet to write down an estimate of $\mathrm { P } ( Z > 6 )$.
Explain how a more reliable estimate of $\mathrm { P } ( Z > 6 )$ could be obtained.
1. State the greatest possible value of $Z$.
2. Explain why it is very unlikely that $Z$ would have this value.
Use the Central Limit Theorem to calculate an estimate of the probability that the mean of 20 independent values of $Z$ is greater than 2 .

OCR MEI Further Statistics B AS 2021 November Q3

3 The weights in kg of male otters in a large river system are known to be Normally distributed with mean 8.3 and standard deviation 1.8. A researcher believes that weights of male otters in another river are higher because of what he suspects is better availability of food. The researcher records the weights of a random sample of 9 male otters in this other river. The sum of these 9 weights is 83.79 kg .

In this question you must show detailed reasoning. You should assume that:
- the weights of otters in the other river are Normally distributed,
- the standard deviation of the weights of otters in the other river is also 1.8 kg .
Show that a test at the $5 \%$ significance level provides sufficient evidence to conclude that the mean weight of male otters in the other river is greater than 8.3 kg .
Explain whether the result of the test suggests that the weights are higher due to better availability of food.
If the standard deviation of the weights of otters in the other river could not be assumed to be 1.8 kg , name an alternative test that the researcher could carry out to investigate otter weights.
Explain why, even if a test at the $5 \%$ significance level results in the rejection of the null hypothesis, you cannot be sure that the alternative hypothesis is true.

OCR MEI Further Statistics B AS 2021 November Q4

4 John regularly downloads podcasts onto his mobile phone. From past experience he knows that the average time to download one 30 -minute podcast is 12.7 s . He believes that this time has recently increased. At each of 12 randomly chosen times, he downloads a 30-minute podcast. The times in seconds to download the 12 podcasts are as follows.
$\begin{array} { l l l l l l l l l l l } 12.63 & 13.24 & 11.73 & 14.91 & 13.17 & 13.53 & 12.33 & 14.27 & 11.48 & 13.51 & 13.05 \end{array} 13.83$

Given that it is not known whether the times are Normally distributed, carry out a suitable test at the $5 \%$ significance level to investigate whether the average download time has increased.
What assumption is required to carry out the test in part (a)?

OCR MEI Further Statistics B AS 2021 November Q5

5 A food company makes mini apple pies. The weight of pastry in a pie is Normally distributed with mean 75 g and standard deviation 4 g . The weight of filling in a pie is Normally distributed with mean 130 g and standard deviation 8 g . You should assume that the weights of pastry and filling in a pie are independent.

Find the probability that the weight of pastry in a randomly chosen pie is between 70 g and 80 g .
Find the probability that the mean weight of filling in 10 randomly chosen pies is at least 125 g. The pies are sold in packs of 4 . The weight of the packaging is Normally distributed with mean 165 g and standard deviation 6 g .
In order to find the probability that the total weight of a pack of 4 pies is less than 1 kg , you must assume that the weight of the packaging is independent of the weight of the pies.
1. State another necessary assumption.
2. Given that the assumptions are valid, calculate this probability.

OCR MEI Further Statistics B AS 2021 November Q6

6 The probability density function of the continuous random variable $X$ is given by
$f ( x ) = \begin{cases} 2 ( 1 + a x ) & 0 \leqslant x \leqslant 1 ,
0 & \text { otherwise } , \end{cases}$
where $a$ is a constant.

Show that $a = - 1$.
Find the cumulative distribution function of $X$.
Find $\mathrm { P } ( X < 0.5 )$.
Show that $\mathrm { E } ( X )$ is greater than the median of $X$.

1	A	B	C	D	E	F	G	H	I	J
1	\(X _ { 1 }\)	\(X _ { 2 }\)	\(X _ { 3 }\)	\(X _ { 4 }\)	\(X _ { 5 }\)	\(X _ { 6 }\)	\(X _ { 7 }\)	\(X _ { 8 }\)	\(Y\)
2	6.31	2.45	3.27	3.06	4.16	1.53	0.43	7.99	3.65
3	1.70	1.52	7.10	8.93	6.44	2.70	9.96	7.83	5.77
4	9.15	0.52	4.95	6.99	6.52	3.15	0.81	5.35	4.68
5	0.65	2.71	7.92	9.65	0.50	4.87	6.46	2.67	4.43
6	3.09	6.11	3.96	0.09	0.18	4.67	0.67	6.20	3.12
7	7.06	5.84	1.97	3.60	9.36	1.97	4.48	3.47	4.72
8	1.46	1.57	5.45	0.37	3.76	7.56	8.48	9.12	4.72
9	9.42	1.85	4.91	1.61	1.94	8.00	1.77	5.34	4.36
10	2.98	5.32	2.91	4.12	9.16	1.76	9.97	6.88	5.39
11	2.83	3.44	3.28	7.85	1.00	0.93	8.77	4.03	4.01
12	4.51	0.59	5.84	9.87	8.65	3.94	7.18	0.23	5.10
13	4.49	0.69	3.65	8.78	4.96	8.96	3.77	1.43	4.59
14	6.57	8.08	4.85	6.75	7.92	0.27	9.69	4.04	6.02
15	8.35	1.09	8.63	8.04	7.23	2.12	2.57	9.59	5.95
16	5.24	9.53	6.08	8.21	3.61	7.07	6.65	7.63	6.75
17	7.89	5.50	3.09	0.71	6.47	5.49	6.47	4.95	5.07
18	8.36	7.27	2.35	9.04	0.58	2.26	3.01	7.90	5.10
19	3.76	1.01	9.61	9.65	7.89	9.98	6.28	4.34	6.56
20	9.94	6.84	3.38	5.53	0.26	8.53	5.72	5.12	5.66
21	7.25	9.10	0.34	2.88	4.66	2.65	6.37	7.63	5.11
22	7.18	7.14	5.38	0.04	4.09	6.47	4.96	4.23	4.94
23	8.69	5.04	4.90	2.94	2.00	4.23	4.13	0.97	4.11
24	3.46	6.33	0.48	9.35	0.23	1.18	7.97	6.37	4.42
25	2.37	7.26	7.16	1.24	5.26	2.80	3.55	3.84	4.19
26	2.16	8.30	7.17	3.32	2.96	1.30	9.11	0.31	4.33
27

	A	B	C	D	E	F	G	H
1	First dice	Second dice	Total (Leila's score) \(\boldsymbol { L }\)	Coin 1	Coin 2	Coin 3	Coin 4	Caleb's score \(\boldsymbol { C }\)
2	1	2	3	H	T	T	T	3
3	6	1	7	T	H	T	T	3
4	2	6	8	H	H	T	T	6
5	2	5	7	T	H	H	H	9
6	1	5	6	T	H	T	T	3
7	5	2	7	H	H	H	H	12
8	1	1	2	H	T	H	T	6
9	2	6	8	T	H	T	H	6
10	6	2	8	H	T	H	T	6
11	1	3	4	T	H	H	H	9
12	6	1	7	T	H	T	T	3
13	3	1	4	T	T	T	T	0
14	3	6	9	H	T	H	H	9
15	2	3	5	T	H	H	H	9
16	2	5	7	H	H	H	H	12
17	1	5	6	H	H	T	H	9
18	5	6	11	T	H	H	H	9
19	4	2	6	T	H	H	T	6
20	6	5	11	T	T	H	H	6
21	1	1	2	T	T	T	T	0

	A	B	C	D
1	\(\mathrm { X } _ { 1 }\)	\(\mathrm { X } _ { 2 }\)	\(\mathrm { X } _ { 3 }\)	T
2	10.9	21.5	5.3	37.7
3	23.9	52.4	85.3	161.6
4	5.2	10.4	24.0	39.6
5	2.9	14.4	0.8	18.1
6	9.0	43.3	49.7	102.0
7	0.4	16.2	12.4	29.0
8	44.1	39.5	22.1	105.7
9	9.2	43.6	13.9	66.7
10	40.4	10.9	6.1	57.4
11	3.2	54.8	15.7	73.7
12	5.3	6.1	1.6	13.0
13	20.5	28.9	22.9	72.3
14	37.3	2.1	28.6	68.0
15	7.1	13.6	50.1	70.8
16	18.6	2.0	9.3	29.9
17	9.0	1.2	49.9	60.1
18	1.9	9.5	69.8	81.2
19	9.0	2.1	10.4	21.5
20	28.7	1.4	93.8	123.9
21	1.8	2.9	34.8	39.5

1	A	B	C	D	E
1	X	Y	\(\mathbf { Z } = \mathbf { 5 } \mathbf { X } - \mathbf { Y }\)	\(\mathbf { Z } > \mathbf { 6 }\)
2	4	13	7	Y
3	4	17	3	N
4	3	21	-6	N
5
6
98	1	14	-9	N
99	5	12	13	Y
100	3	18	-3	N
101	3	15	0	N
102
103			Number of Y	19
104			Number of N	81
105

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