2.04f Find normal probabilities: Z transformation

3 The lifetime of a particular type of light bulb is \(X\) hours, where \(X\) is Normally distributed with mean 1100 and variance 2000.

1. Find \(\mathrm { P } ( 1100 < X < 1200 )\).
2. Use a suitable approximating distribution to find the probability that, in a random sample of 100 of these light bulbs, no more than 40 have a lifetime between 1100 and 1200 hours.
3. A factory has a large number of these light bulbs installed. As soon as \(1 \%\) of the bulbs have come to the end of their lifetimes, it is company policy to replace all of the bulbs. After how many hours should the bulbs need to be replaced?
4. The bulbs are to be replaced by low-energy bulbs. The lifetime of these bulbs is Normally distributed and the mean is claimed by the manufacturer to be 7000 hours. The standard deviation is known to be 100 hours. A random sample of 25 low-energy bulbs is selected. Their mean lifetime is found to be 6972 hours. Carry out a 2 -tail test at the \(10 \%\) level to investigate the claim.
   [0pt] [Question 4 is printed overleaf.]

3 The amount of data, \(X\) megabytes, arriving at an internet server per second during the afternoon is modelled by the Normal distribution with mean 435 and standard deviation 30.

1. Find
   (A) \(\mathrm { P } ( X < 450 )\),
   (B) \(\mathrm { P } ( 400 < X < 450 )\).
2. Find the probability that, during 5 randomly selected seconds, the amounts of data arriving are all between 400 and 450 megabytes. The amount of data, \(Y\) megabytes, arriving at the server during the evening is modelled by the Normal distribution with mean \(\mu\) and standard deviation \(\sigma\).
3. Given that \(\mathrm { P } ( Y < 350 ) = 0.2\) and \(\mathrm { P } ( Y > 390 ) = 0.1\), find the values of \(\mu\) and \(\sigma\).
4. Find values of \(a\) and \(b\) for which \(\mathrm { P } ( a < Y < b ) = 0.95\).

3 Intensity of light is measured in lumens. The random variable \(X\) represents the intensity of the light from a standard 100 watt light bulb. \(X\) is Normally distributed with mean 1720 and standard deviation 90. You may assume that the intensities for different bulbs are independent.

1. Show that \(\mathrm { P } ( X < 1700 ) = 0.4121\).
2. These bulbs are sold in packs of 4 . Find the probability that the intensities of exactly 2 of the 4 bulbs in a randomly chosen pack are below 1700 lumens.
3. Use a suitable approximating distribution to find the probability that the intensities of at least 20 out of 40 randomly selected bulbs are below 1700 lumens. A manufacturer claims that the average intensity of its 25 watt low energy light bulbs is 1720 lumens. A consumer organisation suspects that the true figure may be lower than this. The intensities of a random sample of 20 of these bulbs are measured. A hypothesis test is then carried out to check the claim.
4. Write down a suitable null hypothesis and explain briefly why the alternative hypothesis should be \(\mathrm { H } _ { 1 } : \mu < 1720\). State the meaning of \(\mu\).
5. Given that the standard deviation of the intensity of such bulbs is 90 lumens and that the mean intensity of the sample of 20 bulbs is 1703 lumens, carry out the test at the \(5 \%\) significance level.

3 In a men's cycling time trial, the times are modelled by the random variable \(X\) minutes which is Normally distributed with mean 63 and standard deviation 5.2.

1. Find $$\begin{aligned} & \text { (A) } \mathrm { P } ( X < 65 ) \text {, } \\ & \text { (B) } \mathrm { P } ( 60 < X < 65 ) \text {. } \end{aligned}$$
2. Find the probability that 5 riders selected at random all record times between 60 and 65 minutes.
3. A competitor aims to be in the fastest \(5 \%\) of entrants (i.e. those with the lowest times). Find the maximum time that he can take. It is suggested that holding the time trial on a new course may result in lower times. To investigate this, a random sample of 15 competitors is selected. These 15 competitors do the time trial on the new course. The mean time taken by these riders is 61.7 minutes. You may assume that times are Normally distributed and the standard deviation is still 5.2 minutes. A hypothesis test is carried out to investigate whether times on the new course are lower.
4. Write down suitable null and alternative hypotheses for the test. Carry out the test at the 5\% significance level.

3 The weights of Braeburn apples on display in a supermarket, measured in grams, are Normally distributed with mean 210.5 and standard deviation 15.2.

1. Find the probability that a randomly selected apple weighs at least 220 grams.
2. These apples are sold in packs of 3. You may assume that the weights of apples in each pack are independent. Find the probability that all 3 of the apples in a randomly selected pack weigh at least 220 grams.
3. 100 packs are selected at random.
   (A) State the exact distribution of the number of these 100 packs in which all 3 apples weigh at least 220 grams.
   (B) Use a suitable approximating distribution to find the probability that in at most one of these packs all 3 apples weigh at least 220 grams.
   (C) Explain why this approximating distribution is suitable.
4. The supermarket also sells Cox's Orange Pippin apples. The weights of these apples, measured in grams, are Normally distributed with mean 185 and standard deviation \(\sigma\).
   (A) Given that \(10 \%\) of randomly selected Cox's Orange Pippin apples weigh less than 170 grams, calculate the value of \(\sigma\).
   (B) Sketch the distributions of the weights of both types of apple on a single diagram.

3 At a vineyard, the process used to fill bottles with wine is subject to variation. The contents of bottles are independently Normally distributed with mean \(\mu = 751.4 \mathrm { ml }\) and standard deviation \(\sigma = 2.5 \mathrm { ml }\).

1. Find the probability that a randomly selected bottle contains at least 750 ml .
2. A case of wine consists of 6 bottles. Find the probability that all 6 bottles in a case contain at least 750 ml .
3. Find the probability that, in a random sample of 25 cases, there are at least 2 cases in which all 6 bottles contain at least 750 ml . It is decided to increase the proportion of bottles which contain at least 750 ml to \(98 \%\).
4. This can be done by changing the value of \(\mu\), but retaining the original value of \(\sigma\). Find the required value of \(\mu\).
5. An alternative is to change the value of \(\sigma\), but retain the original value of \(\mu\). Find the required value of \(\sigma\).
6. Comment briefly on which method might be easier to implement and which might be preferable to the vineyard owners.

3 The scores, \(X\), in Paper 1 of an English examination have an underlying Normal distribution with mean 76 and standard deviation 12. The scores are reported as integer marks. So, for example, a score for which \(75.5 \leqslant X < 76.5\) is reported as 76 marks.

1. Find the probability that a candidate's reported mark is 76 .
2. Find the probability that a candidate's reported mark is at least 80 .
3. Three candidates are chosen at random. Find the probability that exactly one of these three candidates' reported marks is at least 80 . The proportion of candidates who receive an A* grade (the highest grade) must not exceed \(10 \%\) but should be as close as possible to \(10 \%\).
4. Find the lowest reported mark that should be awarded an A* grade. The scores in Paper 2 of the examination have an underlying Normal distribution with mean \(\mu\) and standard deviation 12.
5. Given that \(20 \%\) of candidates receive a reported mark of 50 or less, find the value of \(\mu\).

3 The wing lengths of native English male blackbirds, measured in mm , are Normally distributed with mean 130.5 and variance 11.84.

1. Find the probability that a randomly selected native English male blackbird has a wing length greater than 135 mm .
2. Given that \(1 \%\) of native English male blackbirds have wing length more than \(k \mathrm {~mm}\), find the value of \(k\).
3. Find the probability that a randomly selected native English male blackbird has a wing length which is 131 mm correct to the nearest millimetre. It is suspected that Scandinavian male blackbirds have, on average, longer wings than native English male blackbirds. A random sample of 20 Scandinavian male blackbirds has mean wing length 132.4 mm . You may assume that wing lengths in this population are Normally distributed with variance \(11.84 \mathrm {~mm} ^ { 2 }\).
4. Carry out an appropriate hypothesis test, at the \(5 \%\) significance level.
5. Discuss briefly one advantage and one disadvantage of using a \(10 \%\) significance level rather than a \(5 \%\) significance level in hypothesis testing in general.

3 The random variable \(X\) represents the weight in kg of a randomly selected male dog of a particular breed. \(X\) is Normally distributed with mean 30.7 and standard deviation 3.5.

1. Find
   (A) \(\mathrm { P } ( X < 30 )\),
   (B) \(P ( 25 < X < 35 )\).
2. Five of these dogs are chosen at random. Find the probability that each of them weighs at least 30 kg .
3. The weights of females of the same breed of dog are Normally distributed with mean 26.8 kg . Given that \(5 \%\) of female dogs of this breed weigh more than 30 kg , find the standard deviation of their weights.
4. Sketch the distributions of the weights of male and female dogs of this breed on a single diagram.

3 Many types of computer have cooling fans. The random variable \(X\) represents the lifetime in hours of a particular model of cooling fan. \(X\) is Normally distributed with mean 50600 and standard deviation 3400.

1. Find \(\mathrm { P } ( 50000 < X < 55000 )\).
2. The manufacturers claim that at least \(95 \%\) of these fans last longer than 45000 hours. Is this claim valid?
3. Find the value of \(h\) for which \(99.9 \%\) of these fans last \(h\) hours or more.
4. The random variable \(Y\) represents the lifetime in hours of a different model of cooling fan. \(Y\) is Normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that \(\mathrm { P } ( Y < 60000 ) = 0.6\) and \(\mathrm { P } ( Y > 50000 ) = 0.9\). Find the values of \(\mu\) and \(\sigma\).
5. Sketch the distributions of lifetimes for both types of cooling fan on a single diagram.

7 In 1761, James Short took measurements of the parallax of the sun based on the transit of Venus. The mean and standard deviation of a random sample of 50 of these measurements are 8.592 and 0.7534 respectively, in suitable units.

1. Show that if \(X \sim \mathrm {~N} \left( 8.592,0.7534 ^ { 2 } \right)\), then $$\mathrm { P } ( X \leqslant 8.084 ) = \mathrm { P } ( 8.084 < X \leqslant 8.592 ) = \mathrm { P } ( 8.592 < X \leqslant 9.100 ) = \mathrm { P } ( X > 9.100 ) = 0.25 \text {. }$$ The following table summarises the 50 measurements using these intervals.

   Measurement \(( x )\)	\(x \leqslant 8.084\)	\(8.084 < x \leqslant 8.592\)	\(8.592 < x \leqslant 9.100\)	\(x > 9.100\)
   Frequency	8	22	11	9

2. Carry out a test, at the \(\frac { 1 } { 2 } \%\) significance level, of whether a normal distribution fits the data.
3. Obtain a 99\% confidence interval for the mean of all similar parallax measurements.

1 A machine fills packets of flour whose nominal weights are 500 g . Each of a random sample of 100 packets was weighed and 14 packets weighed less than 500 g . The population proportion of packets that weigh less than 500 g is denoted by \(p\).

1. Calculate an approximate \(95 \%\) confidence interval for \(p\).
2. The weights of the packets, in grams, are normally distributed with mean \(\mu\) and variance 50 . Assuming that \(p = 0.14\), calculate the value of \(\mu\).

4 The weights of a particular variety (A) of tomato are known to be Normally distributed with mean 80 grams and standard deviation 11 grams.

1. Find the probability that a randomly chosen tomato of variety A weighs less than 90 grams. The weights of another variety (B) of tomato are known to be Normally distributed with mean 70 grams. These tomatoes are packed in sixes using packaging that weighs 15 grams.
2. The probability that a randomly chosen pack of 6 tomatoes of variety B , including packaging, weighs less than 450 grams is 0.8463 . Show that the standard deviation of the weight of single tomatoes of variety B is 6 grams, to the nearest gram.
3. Tomatoes of variety A are packed in fives using packaging that weighs 25 grams. Find the probability that the total weight of a randomly chosen pack of variety A is greater than the total weight of a randomly chosen pack of variety B .
4. A new variety (C) of tomato is introduced. The weights, \(c\) grams, of a random sample of 60 of these tomatoes are measured giving the following results. $$\Sigma c = 3126.0 \quad \Sigma c ^ { 2 } = 164223.96$$ Find a \(95 \%\) confidence interval for the true mean weight of these tomatoes.

1 Each month the amount of electricity, measured in kilowatt-hours ( kWh ), used by a particular household is Normally distributed with mean 406 and standard deviation 12.

1. Find the probability that, in a randomly chosen month, less than 420 kWh is used. The charge for electricity used is 14.6 pence per kWh .
2. Write down the distribution of the total charge for the amount of electricity used in any one month. Hence find the probability that, in a randomly chosen month, the total charge is more than \(\pounds 60\).
3. The household receives a bill every three months. Assume that successive months may be regarded as independent of each other. Find the value of \(b\) such that the probability that a randomly chosen bill is less than \(\pounds b\) is 0.99 . In a different household, the amount of electricity used per month was Normally distributed with mean 432 kWh . This household buys a new washing machine that is claimed to be cheaper to run than the old one. Over the next six months the amounts of electricity used, in kWh , are as follows. $$\begin{array} { l l l l l l } 404 & 433 & 420 & 423 & 413 & 440 \end{array}$$
4. Treating this as a random sample, carry out an appropriate test, with a \(5 \%\) significance level, to see if there is any evidence to suggest that the amount of electricity used per month by this household has decreased on average.

3 The masses, in kilograms, of a random sample of 100 chickens on sale in a large supermarket were recorded as follows.

1. Assuming that the first and last classes are the same width as the other classes, calculate an estimate of the sample mean and show that the corresponding estimate of the sample standard deviation is 0.2227 kg . A Normal distribution using the mean and standard deviation found in part (i) is to be fitted to these data. The expected frequencies for the classes are as follows.

   Mass \(( m \mathrm {~kg} )\)	\(m < 1.6\)	\(1.6 \leqslant m < 1.8\)	\(1.8 \leqslant m < 2.0\)	\(2.0 \leqslant m < 2.2\)	\(2.2 \leqslant m < 2.4\)	\(2.4 \leqslant m < 2.6\)	\(2.6 \leqslant m\)
   Expected frequency	2.17	10.92	\(f\)	33.85	19.22	5.13	0.68

2. Use the Normal distribution to find \(f\).
3. Carry out a goodness of fit test of this Normal model using a significance level of 5\%.
4. Discuss the outcome of the test with reference to the contributions to the test statistic and to the possibility of other significance levels.

2 In a particular chain of supermarkets, one brand of pasta shapes is sold in small packets and large packets. Small packets have a mean weight of 505 g and a standard deviation of 11 g . Large packets have a mean weight of 1005 g and a standard deviation of 17 g . It is assumed that the weights of packets are Normally distributed and are independent of each other.

1. Find the probability that a randomly chosen large packet weighs between 995 g and 1020 g .
2. Find the probability that the weights of two randomly chosen small packets differ by less than 25 g .
3. Find the probability that the total weight of two randomly chosen small packets exceeds the weight of a randomly chosen large packet.
4. Find the probability that the weight of one randomly chosen small packet exceeds half the weight of a randomly chosen large packet by at least 5 g .
5. A different brand of pasta shapes is sold in packets of which the weights are assumed to be Normally distributed with standard deviation 14 g . A random sample of 20 packets of this pasta is found to have a mean weight of 246 g . Find a \(95 \%\) confidence interval for the population mean weight of these packets.

3 In the manufacture of child car seats, a resin made up of three ingredients is used. The ingredients are two polymers and an impact modifier. The resin is prepared in batches. Each ingredient is supplied by a separate feeder and the amount supplied to each batch, in kg, is assumed to be Normally distributed with mean and standard deviation as shown in the table below. The three feeders are also assumed to operate independently of each other.

1. Find the probability that, in a randomly chosen batch of resin, there is no more than 2100 kg of polymer 1.
2. Find the probability that, in a randomly chosen batch of resin, the amount of polymer 1 exceeds the amount of polymer 2 by at least 400 kg .
3. Find the value of \(b\) such that the total amount of the ingredients in a randomly chosen batch exceeds \(b \mathrm {~kg} 95 \%\) of the time.
4. Polymer 1 costs \(\pounds 1.20\) per kg, polymer 2 costs \(\pounds 1.30\) per kg and the impact modifier costs \(\pounds 0.80\) per kg. Find the mean and variance of the total cost of a batch of resin.
5. Each batch of resin is used to make a large number of car seats from which a random sample of 50 seats is selected in order that the tensile strength (in suitable units) of the resin can be measured. From one such sample, the \(99 \%\) confidence interval for the true mean tensile strength of the resin in that batch was calculated as \(( 123.72,127.38 )\). Find the mean and standard deviation of the sample.

1 Andy, a carpenter, constructs wooden shelf units for storing CDs. The wood used for the shelves has a thickness which is Normally distributed with mean 14 mm and standard deviation 0.55 mm . Andy works to a design which allows a gap of 145 mm between the shelves, but past experience has shown that the gap is Normally distributed with mean 144 mm and standard deviation 0.9 mm . Dimensions of shelves and gaps are assumed to be independent of each other.

1. Find the probability that a randomly chosen gap is less than 145 mm .
2. Find the probability that the combined height of a gap and a shelf is more than 160 mm . A complete unit has 7 shelves and 6 gaps.
3. Find the probability that the overall height of a unit lies between 960 mm and 965 mm . Hence find the probability that at least 3 out of 4 randomly chosen units are between 960 mm and 965 mm high.
4. I buy two randomly chosen CD units made by Andy. The probability that the difference in their heights is less than \(h \mathrm {~mm}\) is 0.95 . Find \(h\).

4 The weights of Avonley Blue cheeses made by a small producer are found to be Normally distributed with mean 10 kg and standard deviation 0.4 kg .

1. Find the probability that a randomly chosen cheese weighs less than 9.5 kg . One particular shop orders four Avonley Blue cheeses each week from the producer. From experience, the shopkeeper knows that the weekly demand from customers for Avonley Blue cheese is Normally distributed with mean 35 kg and standard deviation 3.5 kg . In the interests of food hygiene, no cheese is kept by the shopkeeper from one week to the next.
2. Find the probability that, in a randomly chosen week, demand from customers for Avonley Blue will exceed the supply. Following a campaign to promote Avonley Blue cheese, the shopkeeper finds that the weekly demand for it has increased by \(30 \%\) (i.e. the mean and standard deviation are both increased by \(30 \%\) ). Therefore the shopkeeper increases his weekly order by one cheese.
3. Find the probability that, in a randomly chosen week, demand will now exceed supply.
4. Following complaints, the cheese producer decides to check the mean weight of the Avonley Blue cheeses. For a random sample of 12 cheeses, she finds that the mean weight is 9.73 kg . Assuming that the population standard deviation of the weights is still 0.4 kg , find a \(95 \%\) confidence interval for the true mean weight of the cheeses and comment on the result. Explain what is meant by a 95\% confidence interval. RECOGNISING ACHIEVEMENT

6 The lifetime, \(X\) days, of a particular insect is such that \(\log _ { 10 } X\) has a normal distribution with mean 1.5 and standard deviation 0.2. Find the median lifetime. Find also \(\mathrm { P } ( X \geq 50 )\).

8

1. The variable \(X\) has the distribution \(\mathrm { N } ( 20,9 )\).
   1. Find \(\mathrm { P } ( X > 25 )\).
   2. Given that \(\mathrm { P } ( X > a ) = 0.2\), find \(a\).
   3. Find \(b\) such that \(\mathrm { P } ( 20 - b < X < 20 + b ) = 0.5\).
   4. The variable \(Y\) has the distribution \(\mathrm { N } \left( \mu , \frac { \mu ^ { 2 } } { 9 } \right)\). Find \(\mathrm { P } ( Y > 1.5 \mu )\).

9

1. The masses, in grams, of plums of a certain kind have the distribution \(\mathrm { N } ( 55,18 )\).
   1. Find the probability that a plum chosen at random has a mass between 50.0 and 60.0 grams.
   2. The heaviest \(5 \%\) of plums are classified as extra large. Find the minimum mass of extra large plums.
   3. The plums are packed in bags, each containing 10 randomly selected plums. Find the probability that a bag chosen at random has a total mass of less than 530 g .
2. The masses, in grams, of apples of a certain kind have the distribution \(\mathrm { N } \left( 67 , \sigma ^ { 2 } \right)\). It is given that half of the apples have masses between 62 g and 72 g . Determine \(\sigma\).

7

1. The heights of English men aged 25 to 34 are normally distributed with mean 178 cm and standard deviation 8 cm .
   Three English men aged 25 to 34 are chosen at random. Find the probability that all three men have a height less than 194 cm .
2. The diagram shows the distribution of heights of Scottish women aged 25 to 34. \includegraphics[max width=\textwidth, alt={}, center]{f2f45d6c-cfdc-455b-ab08-597b06a69f36-08_585_1477_909_342} The distribution is approximately normal. Use the diagram in the Printed Answer Booklet to estimate the standard deviation of these heights, explaining your method.

8 A market gardener records the masses of a random sample of 100 of this year's crop of plums. The table shows his results.

1. Explain why the normal distribution might be a reasonable model for this distribution. The market gardener models the distribution of masses by \(\mathrm { N } \left( 47.5,10 ^ { 2 } \right)\).
2. Find the number of plums in the sample that this model would predict to have masses in the range:
   1. \(35 \leq m < 45\)
   2. \(m < 25\).
3. Use your answers to parts (b)(i) and (b)(ii) to comment on the suitability of this model. The market gardener plans to use this model to predict the distribution of the masses of next year's crop of plums.
4. Comment on this plan.

2. \begin{figure}[h] \end{figure} The partially completed box plot in Figure 1 shows the distribution of daily mean air temperatures using the data from the large data set for Beijing in 2015 An outlier is defined as a value
more than \(1.5 \times\) IQR below \(Q _ { 1 }\) or
more than \(1.5 \times\) IQR above \(Q _ { 3 }\) The three lowest air temperatures in the data set are \(7.6 ^ { \circ } \mathrm { C } , 8.1 ^ { \circ } \mathrm { C }\) and \(9.1 ^ { \circ } \mathrm { C }\) The highest air temperature in the data set is \(32.5 ^ { \circ } \mathrm { C }\)

1. Complete the box plot in Figure 1 showing clearly any outliers.
2. Using your knowledge of the large data set, suggest from which month the two outliers are likely to have come. Using the data from the large data set, Simon produced the following summary statistics for the daily mean air temperature, \(x ^ { \circ } \mathrm { C }\), for Beijing in 2015 $$n = 184 \quad \sum x = 4153.6 \quad \mathrm {~S} _ { x x } = 4952.906$$
3. Show that, to 3 significant figures, the standard deviation is \(5.19 ^ { \circ } \mathrm { C }\) Simon decides to model the air temperatures with the random variable $$T \sim \mathrm {~N} \left( 22.6,5.19 ^ { 2 } \right)$$
4. Using Simon's model, calculate the 10th to 90th interpercentile range. Simon wants to model another variable from the large data set for Beijing using a normal distribution.
5. State two variables from the large data set for Beijing that are not suitable to be modelled by a normal distribution. Give a reason for each answer. \includegraphics[max width=\textwidth, alt={}, center]{d1eaaae7-c1dc-4aee-ab54-59f35519a7a4-09_473_1813_2161_127}
   (Total for Question 2 is 11 marks)