2.04f Find normal probabilities: Z transformation

3. A woodwork teacher measures the width, \(w \mathrm {~mm}\), of a board. The measured width, \(X \mathrm {~mm}\), is normally distributed with mean \(w \mathrm {~mm}\) and standard deviation 0.5 mm .

1. Find the probability that \(X\) is within 0.6 mm of \(w\). The same board is measured 16 times and the results are recorded.
2. Find the probability that the mean of these results is within 0.3 mm of \(w\). Given that the mean of these 16 measurements is 35.6 mm ,
3. find a \(98 \%\) confidence interval for \(w\).

7. The heights, in cm, of the male employees in a large company follow a normal distribution with mean 177 and standard deviation 5 The heights, in cm, of the female employees follow a normal distribution with mean 163 and standard deviation 4 A male employee and a female employee are chosen at random.

1. Find the probability that the male employee is taller than the female employee. Six male employees and four female employees are chosen at random.
2. Find the probability that their total height is less than 17 m .

3 When an alarm is raised at a market town's fire station, the fire engine cannot leave until at least five fire-fighters arrive at the station. The call-out time, \(X\) minutes, is the time between an alarm being raised and the fire engine leaving the station. The value of \(X\) was recorded on a random sample of 50 occasions. The results are summarised below, where \(\bar { x }\) denotes the sample mean. $$\sum x = 286.5 \quad \sum ( x - \bar { x } ) ^ { 2 } = 45.16$$

1. Find values for the mean and standard deviation of this sample of 50 call-out times.
2. Hence construct a \(99 \%\) confidence interval for the mean call-out time.
3. The fire and rescue service claims that the station's mean call-out time is less than 5 minutes, whereas a parish councillor suggests that it is more than \(6 \frac { 1 } { 2 }\) minutes. Comment on each of these claims.

4 The time, \(x\) seconds, spent by each of a random sample of 100 customers at an automatic teller machine (ATM) is recorded. The times are summarised in the table.

1. Calculate estimates for the mean and standard deviation of the time spent at the ATM by a customer.
2. The mean time spent at the ATM by a random sample of \(\mathbf { 3 6 }\) customers is denoted by \(\bar { Y }\).
   1. State why the distribution of \(\bar { Y }\) is approximately normal.
   2. Write down estimated values for the mean and standard error of \(\bar { Y }\).
   3. Hence estimate the probability that \(\bar { Y }\) is less than \(1 \frac { 1 } { 2 }\) minutes.

7

1. The weight, \(X\) grams, of soup in a carton may be modelled by a normal random variable with mean 406 and standard deviation 4.2. Find the probability that the weight of soup in a carton:
   1. is less than 400 grams;
   2. is between 402.5 grams and 407.5 grams.
2. The weight, \(Y\) grams, of chopped tomatoes in a tin is a normal random variable with mean \(\mu\) and standard deviation \(\sigma\).
   1. Given that \(\mathrm { P } ( Y < 310 ) = 0.975\), explain why: $$310 - \mu = 1.96 \sigma$$
   2. Given that \(\mathrm { P } ( Y < 307.5 ) = 0.86\), find, to two decimal places, values for \(\mu\) and \(\sigma\).
      (4 marks)

1 In large-scale tree-felling operations, a machine cuts down trees, strips off the branches and then cuts the trunks into logs of length \(X\) metres for transporting to a sawmill. It may be assumed that values of \(X\) are normally distributed with mean \(\mu\) and standard deviation 0.16 , where \(\mu\) can be set to a specific value.

1. Given that \(\mu\) is set to 3.3 , determine:
   1. \(\mathrm { P } ( X < 3.5 )\);
   2. \(\mathrm { P } ( X > 3.0 )\);
   3. \(\mathrm { P } ( 3.0 < X < 3.5 )\).
2. The sawmill now requires a batch of logs such that there is a probability of 0.025 that any given log will have a length less than 3.1 metres. Determine, to two decimal places, the new value of \(\mu\).

3 The height, in metres, of adult male African elephants may be assumed to be normally distributed with mean \(\mu\) and standard deviation 0.20 . The heights of a sample of 12 such elephants were measured with the following results, in metres. $$\begin{array} { l l l l l l l l l l l l } 3.37 & 3.45 & 2.93 & 3.42 & 3.49 & 3.67 & 2.96 & 3.57 & 3.36 & 2.89 & 3.22 & 2.91 \end{array}$$

1. Stating a necessary assumption, construct a \(98 \%\) confidence interval for \(\mu\). (6 marks)
2. The mean height of adult male Asian elephants is known to be 2.90 metres. Using your confidence interval, state, with a reason, what can be concluded about the mean heights of adult males in these two types of elephant.

3 UPVC facia board is supplied in lengths labelled as 5 metres. The actual length, \(X\) metres, of a board may be modelled by a normal distribution with a mean of 5.08 and a standard deviation of 0.05 .

1. Determine:
   1. \(\mathrm { P } ( X < 5 )\);
   2. \(\mathrm { P } ( 5 < X < 5.10 )\).
2. Determine the probability that the mean length of a random sample of 4 boards:
   1. exceeds 5.05 metres;
   2. is exactly 5 metres.
3. Assuming that the value of the standard deviation remains unchanged, determine the mean length necessary to ensure that only 1 per cent of boards have lengths less than 5 metres.

5 The times taken by new recruits to complete an assault course may be modelled by a normal distribution with a standard deviation of 8 minutes. A group of 30 new recruits takes a total time of 1620 minutes to complete the course.

1. Calculate the mean time taken by these 30 new recruits.
2. Assuming that the 30 recruits may be considered to be a random sample, construct a \(98 \%\) confidence interval for the mean time taken by new recruits to complete the course.
3. Construct an interval within which approximately \(98 \%\) of the times taken by individual new recruits to complete the course will lie.
4. State where, if at all, in this question you made use of the Central Limit Theorem.

3 The volume, \(X\) litres, of orange juice in a 1-litre carton may be modelled by a normal distribution with unknown mean \(\mu\). The volumes, \(x\) litres, recorded to the nearest 0.01 litre, in a random sample of 100 cartons are shown in the table.

1. For the group ' \(0.98 - 1.00\) ':
   1. show that it has a mid-point of 0.99 litres;
   2. state the minimum and the maximum values of \(x\) that could be included in this group.
2. Calculate, to three decimal places, estimates of the mean and the standard deviation of these 100 volumes.
3. 1. Construct an approximate \(99 \%\) confidence interval for \(\mu\).
   2. State why use of the Central Limit Theorem was not required when calculating this confidence interval.
   3. Give a reason why the confidence interval is approximate rather than exact.
4. Give a reason in support of the claim that:
   1. \(\mu > 1\);
   2. \(\mathrm { P } ( 0.94 < X < 1.16 )\) is approximately 1 .
      \includegraphics[max width=\textwidth, alt={}]{156f9453-ebc6-4406-b5bc-08d1918ebc62-10_2486_1714_221_153}
      \includegraphics[max width=\textwidth, alt={}]{156f9453-ebc6-4406-b5bc-08d1918ebc62-11_2486_1714_221_153}

6 The volume of shampoo, \(V\) millilitres, delivered by a machine into bottles may be modelled by a normal random variable with mean \(\mu\) and standard deviation \(\sigma\).

1. Given that \(\mu = 412\) and \(\sigma = 8\), determine:
   1. \(\mathrm { P } ( V < 400 )\);
   2. \(\mathrm { P } ( V > 420 )\);
   3. \(\mathrm { P } ( V = 410 )\).
2. A new quality control specification requires that the values of \(\mu\) and \(\sigma\) are changed so that $$\mathrm { P } ( V < 400 ) = 0.05 \quad \text { and } \quad \mathrm { P } ( V > 420 ) = 0.01$$
   1. Show, with the aid of a suitable sketch, or otherwise, that $$400 - \mu = - 1.6449 \sigma \quad \text { and } \quad 420 - \mu = 2.3263 \sigma$$
   2. Hence calculate values for \(\mu\) and \(\sigma\).

3 During June 2011, the volume, \(X\) litres, of unleaded petrol purchased per visit at a supermarket's filling station by private-car customers could be modelled by a normal distribution with a mean of 32 and a standard deviation of 10 .

1. Determine:
   1. \(\mathrm { P } ( X < 40 )\);
   2. \(\mathrm { P } ( X > 25 )\);
   3. \(\mathrm { P } ( 25 < X < 40 )\).
2. Given that during June 2011 unleaded petrol cost \(\pounds 1.34\) per litre, calculate the probability that the unleaded petrol bill for a visit during June 2011 by a private-car customer exceeded \(\pounds 65\).
3. Give two reasons, in context, why the model \(\mathrm { N } \left( 32,10 ^ { 2 } \right)\) is unlikely to be valid for a visit by any customer purchasing fuel at this filling station during June 2011.
   (2 marks)

2 The volume of Everwhite toothpaste in a pump-action dispenser may be modelled by a normal distribution with a mean of 106 ml and a standard deviation of 2.5 ml . Determine the probability that the volume of Everwhite in a randomly selected dispenser is:

1. less than 110 ml ;
2. more than 100 ml ;
3. between 104 ml and 108 ml ;
4. not exactly 106 ml .

6

1. The length of one-metre galvanised-steel straps used in house building may be modelled by a normal distribution with a mean of 1005 mm and a standard deviation of 15 mm . The straps are supplied to house builders in packs of 12, and the straps in a pack may be assumed to be a random sample. Determine the probability that the mean length of straps in a pack is less than one metre.
2. Tania, a purchasing officer for a nationwide house builder, measures the thickness, \(x\) millimetres, of each of a random sample of 24 galvanised-steel straps supplied by a manufacturer. She then calculates correctly that the value of \(\bar { x }\) is 4.65 mm .
   1. Assuming that the thickness, \(X \mathrm {~mm}\), of such a strap may be modelled by the distribution \(\mathrm { N } \left( \mu , 0.15 ^ { 2 } \right)\), construct a \(99 \%\) confidence interval for \(\mu\).
   2. Hence comment on the manufacturer's specification that the mean thickness of such straps is greater than 4.5 mm .

7 A machine, which cuts bread dough for loaves, can be adjusted to cut dough to any specified set weight. For any set weight, \(\mu\) grams, the actual weights of cut dough are known to be approximately normally distributed with a mean of \(\mu\) grams and a fixed standard deviation of \(\sigma\) grams. It is also known that the machine cuts dough to within 10 grams of any set weight.

1. Estimate, with justification, a value for \(\sigma\).
2. The machine is set to cut dough to a weight of 415 grams. As a training exercise, Sunita, the quality control manager, asked Dev, a recently employed trainee, to record the weight of each of a random sample of 15 such pieces of dough selected from the machine's output. She then asked him to calculate the mean and the standard deviation of his 15 recorded weights. Dev subsequently reported to Sunita that, for his sample, the mean was 391 grams and the standard deviation was 95.5 grams. Advise Sunita on whether or not each of Dev's values is likely to be correct. Give numerical support for your answers.
3. Maria, an experienced quality control officer, recorded the weight, \(y\) grams, of each of a random sample of 10 pieces of dough selected from the machine's output when it was set to cut dough to a weight of 820 grams. Her summarised results were as follows. $$\sum y = 8210.0 \quad \text { and } \quad \sum ( y - \bar { y } ) ^ { 2 } = 110.00$$ Explain, with numerical justifications, why both of these values are likely to be correct.

7

1. Electra is employed by E \& G Ltd to install electricity meters in new houses on an estate. Her time, \(X\) minutes, to install a meter may be assumed to be normally distributed with a mean of 48 and a standard deviation of 20 . Determine:
   1. \(\mathrm { P } ( X < 60 )\);
   2. \(\mathrm { P } ( 30 < X < 60 )\);
   3. the time, \(k\) minutes, such that \(\mathrm { P } ( X < k ) = 0.9\).
2. Gazali is employed by E \& G Ltd to install gas meters in the same new houses. His time, \(Y\) minutes, to install a meter has a mean of 37 and a standard deviation of 25 .
   1. Explain why \(Y\) is unlikely to be normally distributed.
   2. State why \(\bar { Y }\), the mean of a random sample of 35 gas meter installations, is likely to be approximately normally distributed.
   3. Determine \(\mathrm { P } ( \bar { Y } > 40 )\).

5 When a particular make of tennis ball is dropped from a vertical distance of 250 cm on to concrete, the height, \(X\) centimetres, to which it first bounces may be assumed to be normally distributed with a mean of 140 and a standard deviation of 2.5.

1. Determine:
   1. \(\mathrm { P } ( X < 145 )\);
   2. \(\mathrm { P } ( 138 < X < 142 )\).
2. Determine, to one decimal place, the maximum height exceeded by \(85 \%\) of first bounces.
3. Determine the probability that, for a random sample of 4 first bounces, the mean height is greater than 139 cm .

3 The weight, \(X\) grams, of talcum powder in a tin may be modelled by a normal distribution with mean 253 and standard deviation \(\sigma\).

1. Given that \(\sigma = 5\), determine:
   1. \(\mathrm { P } ( X < 250 )\);
   2. \(\mathrm { P } ( 245 < X < 250 )\);
   3. \(\mathrm { P } ( X = 245 )\).
2. Assuming that the value of the mean remains unchanged, determine the value of \(\sigma\) necessary to ensure that \(98 \%\) of tins contain more than 245 grams of talcum powder.
   (4 marks) \includegraphics[max width=\textwidth, alt={}, center]{adf1c0d2-b0a6-4a2f-baf2-cfb45d771315-07_38_118_440_159} \includegraphics[max width=\textwidth, alt={}, center]{adf1c0d2-b0a6-4a2f-baf2-cfb45d771315-07_40_118_529_159}

3 Each day, Margot completes the crossword in her local morning newspaper. Her completion times, \(X\) minutes, can be modelled by a normal random variable with a mean of 65 and a standard deviation of 20 .

1. Determine:
   1. \(\mathrm { P } ( X < 90 )\);
   2. \(\mathrm { P } ( X > 60 )\).
2. Given that Margot's completion times are independent from day to day, determine the probability that, during a particular period of 6 days:
   1. she completes one of the six crosswords in exactly 60 minutes;
   2. she completes each crossword in less than 60 minutes;
   3. her mean completion time is less than 60 minutes.
      
      \includegraphics[max width=\textwidth, alt={}]{c4844a30-6a86-49e3-b6aa-8e213dfc8ca1-07_2484_1709_223_153}

2 The diameter, \(D\) millimetres, of an American pool ball may be modelled by a normal random variable with mean 57.15 and standard deviation 0.04 .

1. Determine:
   1. \(\mathrm { P } ( D < 57.2 )\);
   2. \(\mathrm { P } ( 57.1 < D < 57.2 )\).
2. A box contains 16 of these pool balls. Given that the balls may be regarded as a random sample, determine the probability that:
   1. all 16 balls have diameters less than 57.2 mm ;
   2. the mean diameter of the 16 balls is greater than 57.16 mm .

5 A general store sells lawn fertiliser in 2.5 kg bags, 5 kg bags and 10 kg bags.

1. The actual weight, \(W\) kilograms, of fertiliser in a 2.5 kg bag may be modelled by a normal random variable with mean 2.75 and standard deviation 0.15 . Determine the probability that the weight of fertiliser in a 2.5 kg bag is:
   1. less than 2.8 kg ;
   2. more than 2.5 kg .
2. The actual weight, \(X\) kilograms, of fertiliser in a 5 kg bag may be modelled by a normal random variable with mean 5.25 and standard deviation 0.20 .
   1. Show that \(\mathrm { P } ( 5.1 < X < 5.3 ) = 0.372\), correct to three decimal places.
   2. A random sample of four 5 kg bags is selected. Calculate the probability that none of the four bags contains between 5.1 kg and 5.3 kg of fertiliser.
3. The actual weight, \(Y\) kilograms, of fertiliser in a 10 kg bag may be modelled by a normal random variable with mean 10.75 and standard deviation 0.50. A random sample of six 10 kg bags is selected. Calculate the probability that the mean weight of fertiliser in the six bags is less than 10.5 kg .

2 The weight, \(X\) grams, of the contents of a tin of baked beans can be modelled by a normal random variable with a mean of 421 and a standard deviation of 2.5.

1. Find:
   1. \(\mathrm { P } ( X = 421 )\);
   2. \(\mathrm { P } ( X < 425 )\);
   3. \(\mathrm { P } ( 418 < X < 424 )\).
2. Determine the value of \(x\) such that \(\mathrm { P } ( X < x ) = 0.98\).
3. The weight, \(Y\) grams, of the contents of a tin of ravioli can be modelled by a normal random variable with a mean of \(\mu\) and a standard deviation of 3.0 . Find the value of \(\mu\) such that \(\mathrm { P } ( Y < 410 ) = 0.01\).

6 The weight, \(X\) kilograms, of sand in a bag can be modelled by a normal random variable with unknown mean \(\mu\) and known standard deviation 0.4 .

1. The sand in each of a random sample of 25 bags from a large batch is weighed. The total weight of sand in these 25 bags is found to be 497.5 kg .
   1. Construct a 98\% confidence interval for the mean weight of sand in bags in the batch.
   2. Hence comment on the claim that bags in the batch contain an average of 20 kg of sand.
   3. State why use of the Central Limit Theorem is not required in answering part (a)(i).
2. The weight, \(Y\) kilograms, of cement in a bag can be modelled by a normal random variable with mean 25.25 and standard deviation 0.35. A firm purchases 10 such bags. These bags may be considered to be a random sample.
   1. Determine the probability that the mean weight of cement in the 10 bags is less than 25 kg .
   2. Calculate the probability that the weight of cement in each of the 10 bags is more than 25 kg .
      \includegraphics[max width=\textwidth, alt={}]{fbee7665-54e4-4805-9ce0-6244a4ba043c-20_1111_1707_1592_153}
      \includegraphics[max width=\textwidth, alt={}]{fbee7665-54e4-4805-9ce0-6244a4ba043c-23_2351_1707_219_153}

2

1. Tim rings the church bell in his village every Sunday morning. The time that he spends ringing the bell may be modelled by a normal distribution with mean 7.5 minutes and standard deviation 1.6 minutes. Determine the probability that, on a particular Sunday morning, the time that Tim spends ringing the bell is:
   1. at most 10 minutes;
   2. more than 6 minutes;
   3. between 5 minutes and 10 minutes.
2. June rings the same church bell for weekday weddings. The time that she spends, in minutes, ringing the bell may be modelled by the distribution \(\mathrm { N } \left( \mu , 2.4 ^ { 2 } \right)\). Given that 80 per cent of the times that she spends ringing the bell are less than 15 minutes, find the value of \(\mu\).
   [0pt] [4 marks]
   \includegraphics[max width=\textwidth, alt={}]{ddf7f158-b6ae-42c6-98f1-d59c205646ad-04_1477_1707_1226_153}

2 A garden centre sells bamboo canes of nominal length 1.8 metres. The length, \(X\) metres, of the canes can be modelled by a normal distribution with mean 1.86 and standard deviation \(\sigma\).

1. Assuming that \(\sigma = 0.04\), determine:
   1. \(\mathrm { P } ( X < 1.90 )\);
   2. \(\mathrm { P } ( X > 1.80 )\);
   3. \(\mathrm { P } ( 1.80 < X < 1.90 )\);
   4. \(\mathrm { P } ( X \neq 1.86 )\).
2. It is subsequently found that \(\mathrm { P } ( X > 1.80 ) = 0.98\). Determine the value of \(\sigma\).
   [0pt] [3 marks]
   \includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-06_1529_1717_1178_150}