2.04e Normal distribution: as model N(mu, sigma^2)

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 .
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      \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]
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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}

7 The volume of water, \(V\), used by a guest in an en suite shower room at a small guest house may be modelled by a random variable with mean \(\mu\) litres and standard deviation 65 litres. A random sample of 80 guests using this shower room showed a mean usage of 118 litres of water.

1. 1. Give a numerical justification as to why \(V\) is unlikely to be normally distributed.
   2. Explain why \(\bar { V }\), the mean of a random sample of 80 observations of \(V\), may be assumed to be approximately normally distributed.
2. 1. Construct a \(98 \%\) confidence interval for \(\mu\).
   2. Hence comment on a claim that \(\mu\) is 140 .
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      \includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-25_2484_1707_221_153}
      \includegraphics[max width=\textwidth, alt={}, center]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-27_2490_1719_217_150} \includegraphics[max width=\textwidth, alt={}, center]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-28_2486_1728_221_141}

5 Still mineral water is supplied in 1.5-litre bottles. The actual volume, \(X\) millilitres, in a bottle may be modelled by a normal distribution with mean \(\mu = 1525\) and standard deviation \(\sigma = 9.6\).

1. Determine the probability that the volume of water in a randomly selected bottle is:
   1. less than 1540 ml ;
   2. more than 1535 ml ;
   3. between 1515 ml and 1540 ml ;
   4. not 1500 ml .
2. The supplier requires that only 10 per cent of bottles should contain more than 1535 ml of water. Assuming that there has been no change in the value of \(\sigma\), calculate the reduction in the value of \(\mu\) in order to satisfy this requirement. Give your answer to one decimal place.
3. Sparkling spring water is supplied in packs of six 0.5 -litre bottles. The actual volume in a bottle may be modelled by a normal distribution with mean 508.5 ml and standard deviation 3.5 ml . Stating a necessary assumption, determine the probability that:
   1. the volume of water in each of the 6 bottles from a randomly selected pack is more than 505 ml ;
   2. the mean volume of water in the 6 bottles from a randomly selected pack is more than 505 ml .
      [0pt] [7 marks]

7. The times taken by a large number of people to read a certain book can be modelled by a normal distribution with mean \(5 \cdot 2\) hours. It is found that \(62 \cdot 5 \%\) of the people took more than \(4 \cdot 5\) hours to read the book.

1. Show that the standard deviation of the times is approximately \(2 \cdot 2\) hours.
2. Calculate the percentage of the people who took between 4 and 7 hours to read the book.
3. Calculate the probability that two of the people chosen at random both took less than 5 hours to read the book, stating any assumption that you make.
4. If a number of extra people were taken into account, all of whom took exactly \(5 \cdot 2\) hours to read the book, state with reasons what would happen to (i) the mean, (ii) the variance and explain briefly why the distribution would no longer be normal.

4. A botanist believes that the lengths of the branches on trees of a certain species can be modelled by a normal distribution.
When he measures the lengths of 500 branches, he finds 55 which are less than 30 cm long and 200 which are more than 90 cm long.

1. Find the mean and the standard deviation of the lengths.
2. In a sample of 1000 branches, how many would he expect to find with lengths greater than 1 metre? \section*{STATISTICS 1 (A) TEST PAPER 7 Page 2}

7. The volume of liquid in bottles of sparkling water from one producer is believed to be normally distributed with a mean of 704 ml and a variance of \(3.2 \mathrm { ml } ^ { 2 }\). Calculate the probability that a randomly chosen bottle from this producer contains

1. more than 706 ml ,
2. between 703 and 708 ml . The bottles are labelled as containing 700 ml .
3. In a delivery of 1200 bottles, how many could be expected to contain less than the stated 700 ml ? The bottling process can be adjusted so that the mean changes but the variance is unchanged.
4. What should the mean be changed to in order to have only a \(0.1 \%\) chance of a bottle having less than 700 ml of sparkling water? Give your answer correct to 1 decimal place.

5. The time taken in minutes, \(T\), for a mechanic to service a bicycle follows a normal distribution with a mean of 25 minutes and a variance of 16 minutes \(^ { 2 }\). Find

1. \(\mathrm { P } ( T < 28 )\),
2. \(\quad \mathrm { P } ( | T - 25 | < 5 )\). One afternoon the mechanic has 3 bicycles to service.
3. Find the probability that he will take less than 23 minutes on each of the three bicycles.
   (4 marks)

3. The random variable \(X\) is normally distributed with a mean of 42 and a variance of 18 . Find

1. \(\mathrm { P } ( X \leq 45 )\),
2. \(\mathrm { P } ( 32 \leq X \leq 38 )\),
3. the value of \(x\) such that \(\mathrm { P } ( X \leq x ) = 0.95\)

3. The time it takes girls aged 15 to complete an obstacle course is found to be normally distributed with a mean of 21.5 minutes and a standard deviation of 2.2 minutes.

1. Find the probability that a randomly chosen 15 year-old girl completes the course in less than 25 minutes. A 13 year-old girl completes the course in exactly 19 minutes.
2. What percentage of 15 year-old girls would she beat over the course? Anyone completing the course in less than 20 minutes is presented with a certificate of achievement. Three friends all complete the course one afternoon.
3. What is the probability that exactly two of them get certificates?

4. The random variable \(A\) is normally distributed with a mean of 32.5 and a variance of 18.6 Find

1. \(\mathrm { P } ( A < 38.2 )\),
2. \(\mathrm { P } ( 31 \leq A \leq 35 )\), The random variable \(B\) is normally distributed with a standard deviation of 7.2
   Given also that \(\mathrm { P } ( B > 110 ) = 0.138\),
3. find the mean of \(B\).

2.

1. Explain briefly what is meant by a statistical model.
2. State, with a reason, whether or not the normal distribution might be suitable for modelling each of the following:
   1. The number of children in a family;
   2. The time taken for a particular employee to cycle to work each day using the same route;
   3. The quarterly electricity bills for a particular house.

4. A company produces jars of English Honey. The weight of the glass jars used is normally distributed with a mean of 122.3 g and a standard deviation of 2.6 g . Calculate the probability that a randomly chosen jar will weigh

1. less than 127 g ,
2. less than 121.5 g . The weight of honey put into each jar by a machine is normally distributed with a standard deviation of 1.6 g . The machine operator can adjust the mean weight of the honey put into each jar without changing the standard deviation.
3. Find, correct to 4 significant figures, the minimum that the mean weight can be set to such that at most 1 in 20 of the jars will contain less than 454 g .
   (4 marks)

3. A study was made of the heights of boys of different ages in Lancashire. The study concluded that the heights of 13 year-old boys are normally distributed with a mean of 156 cm and a variance of \(73 \mathrm {~cm} ^ { 2 }\). Find the probability that a 13 year-old boy chosen at random will be

1. more than 165 cm tall,
2. between 156 and 165 cm tall. The study also concluded that the heights of 14 year-old boys are normally distributed with a mean of 160 cm and a variance of \(79 \mathrm {~cm} ^ { 2 }\). One 13 year-old and one 14 year-old boy are chosen at random.
3. Find the probability that both boys are more than 165 cm tall.
4. State, with a reason, whether the probability that the combined height of the two boys is more than 330 cm is more or less than your answer to part (c).
   (2 marks)

4. Alan runs on a treadmill each day for as long as he can at 7 miles per hour. The length of time for which he runs is normally distributed with a mean of 21.6 minutes and a standard deviation of 1.8 minutes.

1. Calculate the probability that on any one day Alan will run for less than 20 minutes.
2. Estimate the number of times in a ninety-day period that Alan will run for more than 24 minutes.
3. On a particular day Alan is still running after 22 minutes. Find the probability that he will stop running in the next 2 minutes.

7. A cyber-cafe recorded how long each user stayed during one day giving the following results.

1. Use linear interpolation to estimate the median and quartiles of these data. The results of a previous study had led to the suggestion that the length of time each user stays can be modelled by a normal distribution with a mean of 72 minutes and a standard deviation of 48 minutes.
2. Find the median and quartiles that this model would predict.
3. Comment on the suitability of the suggested model in the light of the new results.

3. The time that a school pupil spends on French homework each week is normally distributed with a mean of 55 minutes and a standard deviation of 10 minutes. The time that this pupil spends on English homework each week is normally distributed with a mean of 1 hour 30 minutes and a standard deviation of 18 minutes. Find the probability that in a randomly chosen week

1. the pupil spends more than 2 hours in total doing French and English homework,
2. the pupil spends more than twice as long doing English homework as he spends doing French homework.
   (6 marks)

6. The weight of a particular electrical component is normally distributed with a mean of 46.7 grams and a variance of 1.8 grams \(^ { 2 }\). The component is sold in boxes of 12 .

1. State the distribution of the mean weight of the components in one box.
2. Find the probability that the mean weight of the components in a randomly chosen box is more than 47 grams.
   (3 marks)
   After a break in production the component manufacturer wishes to find out if the mean weight of the components has changed. A random sample of 30 components is found to have a mean weight of 46.5 grams.
3. Assuming that the variance of the weight of the components is unchanged, test at the \(5 \%\) level of significance if there has been any change in the mean weight of the components.
   (7 marks)

6. A butter packing machine cuts butter into blocks. The weight of a block of butter is normally distributed with a mean weight of 250 g and a standard deviation of 4 g . A random sample of 15 blocks is taken to monitor any change in the mean weight of the blocks of butter.

1. Find the critical region of a suitable test using a \(2 \%\) level of significance.
   (3)
2. Assuming the mean weight of a block of butter has increased to 254 g , find the probability of a Type II error.

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 \(Y = 0.7 X\).

1. Find the probability that a randomly chosen peeled orange weighs less than 250 grams. I randomly choose 5 oranges to buy.
2. Find the probability that the total weight of the 5 unpeeled oranges is at least 1800 grams.
3. 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.