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

The length of time, \(L\) hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours.

1. Find \(\mathrm { P } ( L > 127 )\).
2. Find the value of \(d\) such that \(\mathrm { P } ( L < d ) = 0.10\) Alice is about to go on a 6 hour journey.
   Given that it is 127 hours since Alice last charged her phone,
3. find the probability that her phone will not need charging before her journey is completed.

1. A survey of 100 households gave the following results for weekly income \(\pounds y\).

(You may use \(\sum f y ^ { 2 } = 12452\) 800)
A histogram was drawn and the class \(200 \leqslant y < 240\) was represented by a rectangle of width 2 cm and height 7 cm .

1. Calculate the width and the height of the rectangle representing the class $$320 \leqslant y < 400$$
2. Use linear interpolation to estimate the median weekly income to the nearest pound.
3. Estimate the mean and the standard deviation of the weekly income for these data. One measure of skewness is \(\frac { 3 ( \text { mean } - \text { median } ) } { \text { standard deviation } }\).
4. Use this measure to calculate the skewness for these data and describe its value. Katie suggests using the random variable \(X\) which has a normal distribution with mean 320 and standard deviation 150 to model the weekly income for these data.
5. Find \(\mathrm { P } ( 240 < X < 400 )\).
6. With reference to your calculations in parts (d) and (e) and the data in the table, comment on Katie's suggestion.

3. The continuous random variable \(Y\) is normally distributed with mean 100 and variance 256 .

1. Find \(\mathrm { P } ( Y < 80 )\).
2. Find \(k\) such that \(\mathrm { P } ( 100 - k \leq Y \leq 100 + k ) = 0.516\).

5. A random variable \(X\) has a normal distribution.

1. Describe two features of the distribution of \(X\). A company produces electronic components which have life spans that are normally distributed. Only \(1 \%\) of the components have a life span less than 3500 hours and \(2.5 \%\) have a life span greater than 5500 hours.
2. Determine the mean and standard deviation of the life spans of the components. The company gives warranty of 4000 hours on the components.
3. Find the proportion of components that the company can expect to replace under the warranty.

5. A health club lets members use, on each visit, its facilities for as long as they wish. The club's records suggest that the length of a visit can be modelled by a normal distribution with mean 90 minutes. Only \(20 \%\) of members stay for more than 125 minutes.

1. Find the standard deviation of the normal distribution.
2. Find the probability that a visit lasts less than 25 minutes. The club introduce a closing time of 10:00 pm. Tara arrives at the club at 8:00 pm.
3. Explain whether or not this normal distribution is still a suitable model for the length of her visit.

1. A scientist found that the time taken, \(M\) minutes, to carry out an experiment can be modelled by a normal random variable with mean 155 minutes and standard deviation 3.5 minutes.

Find

1. \(\mathrm { P } ( M > 160 )\).
2. \(\mathrm { P } ( 150 \leqslant M \leqslant 157 )\).
3. the value of \(m\), to 1 decimal place, such that \(\mathrm { P } ( M \leqslant m ) = 0.30\).
   

5. From experience a high-jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts. Assuming that the heights the high-jumper can reach follow a Normal distribution,

1. draw a sketch to illustrate the above information,
2. find, to 3 decimal places, the mean and the standard deviation of the heights the high-jumper can reach,
3. calculate the probability that he can jump at least 1.74 m .

2. The box plot in Figure 1 shows a summary of the weights of the luggage, in kg, for each musician in an orchestra on an overseas tour. \begin{figure}[h] \end{figure} The airline's recommended weight limit for each musician's luggage was 45 kg . Given that none of the musicians' luggage weighed exactly 45 kg ,

1. state the proportion of the musicians whose luggage was below the recommended weight limit. A quarter of the musicians had to pay a charge for taking heavy luggage.
2. State the smallest weight for which the charge was made.
3. Explain what you understand by the + on the box plot in Figure 1, and suggest an instrument that the owner of this luggage might play.
4. Describe the skewness of this distribution. Give a reason for your answer. One musician of the orchestra suggests that the weights of luggage, in kg, can be modelled by a normal distribution with quartiles as given in Figure 1.
5. Find the standard deviation of this normal distribution.

6. The random variable \(X\) has a normal distribution with mean 20 and standard deviation 4 .

1. Find \(\mathrm { P } ( X > 25 )\).
2. Find the value of \(d\) such that \(\mathrm { P } ( 20 < X < d ) = 0.4641\)

7. A packing plant fills bags with cement. The weight \(X \mathrm {~kg}\) of a bag of cement can be modelled by a normal distribution with mean 50 kg and standard deviation 2 kg .

1. Find \(\mathrm { P } ( X > 53 )\).
2. Find the weight that is exceeded by \(99 \%\) of the bags. Three bags are selected at random.
3. Find the probability that two weigh more than 53 kg and one weighs less than 53 kg .

8. The lifetimes of bulbs used in a lamp are normally distributed. A company \(X\) sells bulbs with a mean lifetime of 850 hours and a standard deviation of 50 hours.

1. Find the probability of a bulb, from company \(X\), having a lifetime of less than 830 hours.
2. In a box of 500 bulbs, from company \(X\), find the expected number having a lifetime of less than 830 hours. A rival company \(Y\) sells bulbs with a mean lifetime of 860 hours and \(20 \%\) of these bulbs have a lifetime of less than 818 hours.
3. Find the standard deviation of the lifetimes of bulbs from company \(Y\). Both companies sell the bulbs for the same price.
4. State which company you would recommend. Give reasons for your answer.

7. The distances travelled to work, \(D \mathrm {~km}\), by the employees at a large company are normally distributed with \(D \sim \mathrm {~N} \left( 30,8 ^ { 2 } \right)\).

1. Find the probability that a randomly selected employee has a journey to work of more than 20 km .
2. Find the upper quartile, \(Q _ { 3 }\), of \(D\).
3. Write down the lower quartile, \(Q _ { 1 }\), of \(D\). An outlier is defined as any value of \(D\) such that \(D < h\) or \(D > k\) where $$h = Q _ { 1 } - 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right) \quad \text { and } \quad k = Q _ { 3 } + 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$$
4. Find the value of \(h\) and the value of \(k\). An employee is selected at random.
5. Find the probability that the distance travelled to work by this employee is an outlier.

The heights of an adult female population are normally distributed with mean 162 cm and standard deviation 7.5 cm .

1. Find the probability that a randomly chosen adult female is taller than 150 cm .
   (3) Sarah is a young girl. She visits her doctor and is told that she is at the 60th percentile for height.
2. Assuming that Sarah remains at the 60th percentile, estimate her height as an adult. The heights of an adult male population are normally distributed with standard deviation 9.0 cm . Given that \(90 \%\) of adult males are taller than the mean height of adult females,
3. find the mean height of an adult male.

The time, in minutes, taken to fly from London to Malaga has a normal distribution with mean 150 minutes and standard deviation 10 minutes.

1. Find the probability that the next flight from London to Malaga takes less than 145 minutes. The time taken to fly from London to Berlin has a normal distribution with mean 100 minutes and standard deviation \(d\) minutes. Given that \(15 \%\) of the flights from London to Berlin take longer than 115 minutes,
2. find the value of the standard deviation \(d\). The time, \(X\) minutes, taken to fly from London to another city has a normal distribution with mean \(\mu\) minutes. Given that \(\mathrm { P } ( X < \mu - 15 ) = 0.35\)
3. find \(\mathrm { P } ( X > \mu + 15 \mid X > \mu - 15 )\).

1. The weight, in grams, of beans in a tin is normally distributed with mean \(\mu\) and standard deviation 7.8

Given that \(10 \%\) of tins contain less than 200 g , find

1. the value of \(\mu\)
2. the percentage of tins that contain more than 225 g of beans. The machine settings are adjusted so that the weight, in grams, of beans in a tin is normally distributed with mean 205 and standard deviation \(\sigma\).
3. Given that \(98 \%\) of tins contain between 200 g and 210 g find the value of \(\sigma\).

1. The discrete random variable \(X\) has probability distribution

$$\mathrm { P } ( X = x ) = \frac { 1 } { 10 } \quad x = 1,2,3 , \ldots 10$$

1. Write down the name given to this distribution.
2. Write down the value of
   1. \(\mathrm { P } ( X = 10 )\)
   2. \(\mathrm { P } ( X < 10 )\) The continuous random variable \(Y\) has the normal distribution \(\mathrm { N } \left( 10,2 ^ { 2 } \right)\)
3. Write down the value of
   1. \(\mathrm { P } ( Y = 10 )\)
   2. \(\mathrm { P } ( Y < 10 )\)

6. The time taken, in minutes, by children to complete a mathematical puzzle is assumed to be normally distributed with mean \(\mu\) and standard deviation \(\sigma\). The puzzle can be completed in less than 24 minutes by \(80 \%\) of the children. For \(5 \%\) of the children it takes more than 28 minutes to complete the puzzle.

1. Show this information on the Normal curve below.
2. Write down the percentage of children who take between 24 minutes and 28 minutes to complete the puzzle.
3. 1. Find two equations in \(\mu\) and \(\sigma\).
   2. Hence find, to 3 significant figures, the value of \(\mu\) and the value of \(\sigma\). A child is selected at random.
4. Find the probability that the child takes less than 12 minutes to complete the puzzle. \includegraphics[max width=\textwidth, alt={}, center]{ca8418eb-4d35-40f4-af40-77503327ae52-11_314_1255_1375_356}

7. The heights of adult females are normally distributed with mean 160 cm and standard deviation 8 cm .

1. Find the probability that a randomly selected adult female has a height greater than 170 cm . Any adult female whose height is greater than 170 cm is defined as tall. An adult female is chosen at random. Given that she is tall,
2. find the probability that she has a height greater than 180 cm . Half of tall adult females have a height greater than \(h \mathrm {~cm}\).
3. Find the value of \(h\).

The random variable \(Z \sim \mathrm {~N} ( 0,1 )\) \(A\) is the event \(Z > 1.1\) \(B\) is the event \(Z > - 1.9\) \(C\) is the event \(- 1.5 < Z < 1.5\)

1. Find
   1. \(\mathrm { P } ( A )\)
   2. \(\mathrm { P } ( B )\)
   3. \(\mathrm { P } ( C )\)
   4. \(\mathrm { P } ( A \cup C )\) The random variable \(X\) has a normal distribution with mean 21 and standard deviation 5
2. Find the value of \(w\) such that \(\mathrm { P } ( X > w \mid X > 28 ) = 0.625\)

5. A midwife records the weights, in kg , of a sample of 50 babies born at a hospital. Her results are given in the table below. [You may use \(\sum \mathrm { f } x ^ { 2 } = 611.375\) ] A histogram has been drawn to represent these data. The bar representing the weight \(2 \leqslant w < 3\) has a width of 1 cm and a height of 4 cm .

1. Calculate the width and height of the bar representing a weight of \(3 \leqslant w < 3.5\)
2. Use linear interpolation to estimate the median weight of these babies.
3. 1. Show that an estimate of the mean weight of these babies is 3.43 kg .
   2. Find an estimate of the standard deviation of the weights of these babies. Shyam decides to model the weights of babies born at the hospital, by the random variable \(W\), where \(W \sim \mathrm {~N} \left( 3.43,0.65 ^ { 2 } \right)\)
4. Find \(\mathrm { P } ( W < 3 )\)
5. With reference to your answers to (b), (c)(i) and (d) comment on Shyam's decision. A newborn baby weighing 3.43 kg is born at the hospital.
6. Without carrying out any further calculations, state, giving a reason, what effect the addition of this newborn baby to the sample would have on your estimate of the
   1. mean,
   2. standard deviation.

6. The time, in minutes, taken by men to run a marathon is modelled by a normal distribution with mean 240 minutes and standard deviation 40 minutes.

1. Find the proportion of men that take longer than 300 minutes to run a marathon.
   (3) Nathaniel is preparing to run a marathon. He aims to finish in the first 20\% of male runners.
2. Using the above model estimate the longest time that Nathaniel can take to run the marathon and achieve his aim.
   (3) The time, \(W\) minutes, taken by women to run a marathon is modelled by a normal distribution with mean \(\mu\) minutes. Given that \(\mathrm { P } ( W < \mu + 30 ) = 0.82\)
3. find \(\mathrm { P } ( W < \mu - 30 \mid W < \mu )\)

2. An estate agent is studying the cost of office space in London. He takes a random sample of 90 offices and calculates the cost, \(\pounds x\) per square foot. His results are given in the table below. A histogram is drawn for these data and the bar representing \(50 \leqslant x < 60\) is 2 cm wide and 8 cm high.

1. Calculate the width and height of the bar representing \(20 \leqslant x < 40\)
2. Use linear interpolation to estimate the median cost.
3. Estimate the mean cost of office space for these data.
4. Estimate the standard deviation for these data.
5. Describe, giving a reason, the skewness. Rika suggests that the cost of office space in London can be modelled by a normal distribution with mean \(\pounds 50\) and standard deviation \(\pounds 10\)
6. With reference to your answer to part (e), comment on Rika's suggestion.
7. Use Rika's model to estimate the 80th percentile of the cost of office space in London.

5. Yuto works in the quality control department of a large company. The time, \(T\) minutes, it takes Yuto to analyse a sample is normally distributed with mean 18 minutes and standard deviation 5 minutes.

1. Find the probability that Yuto takes longer than 20 minutes to analyse the next sample. (3) The company has a large store of samples analysed by Yuto with the time taken for each analysis recorded. Serena is investigating the samples that took Yuto longer than 15 minutes to analyse. She selects, at random, one of the samples that took Yuto longer than 15 minutes to analyse.
2. Find the probability that this sample took Yuto more than 20 minutes to analyse. Serena can identify, in advance, the samples that Yuto can analyse in under 15 minutes and in future she will assign these to someone else.
3. Estimate the median time taken by Yuto to analyse samples in future.

3. The random variable \(Y\) has a normal distribution with mean \(\mu\) and standard deviation \(\sigma\) The \(\mathrm { P } ( Y > 17 ) = 0.4\) Find

1. \(\mathrm { P } ( \mu < Y < 17 )\)
2. \(\mathrm { P } ( \mu - \sigma < Y < 17 )\)

7. Farmer Adam grows potatoes. The weights of potatoes, in grams, grown by Adam are normally distributed with a mean of 140 g and a standard deviation of 40 g . Adam cannot sell potatoes with a weight of less than 92 g .

1. Find the percentage of potatoes that Adam grows but cannot sell. The upper quartile of the weight of potatoes sold by Adam is \(q _ { 3 }\)
2. Find the probability that the weight of a randomly selected potato grown by Adam is more than \(q _ { 3 }\)
3. Find the lower quartile, \(q _ { 1 }\), of the weight of potatoes sold by Adam. Betty selects a random sample of 3 potatoes sold by Adam.
4. Find the probability that one weighs less than \(q _ { 1 }\), one weighs more than \(q _ { 3 }\) and one has a weight between \(q _ { 1 }\) and \(q _ { 3 }\)

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