Justifying CLT for sampling distribution

7 Previous records have shown that the number of cars entering Bampor on any day has mean 352 and variance 121.

1. Find the probability that the mean number of cars entering Bampor during a random sample of 200 days is more than 354 .
2. State, with a reason, whether it was necessary to assume that the number of cars entering Bampor on any day has a normal distribution in order to find the probability in part (i).
3. It is thought that the population mean may recently have changed. The number of cars entering Bampor during the day was recorded for each of a random sample of 50 days and the sample mean was found to be 356 . Assuming that the variance is unchanged, test at the \(5 \%\) significance level whether the population mean is still 352 .

5 The mass, in kilograms, of rocks in a certain area has mean 14.2 and standard deviation 3.1.

1. Find the probability that the mean mass of a random sample of 50 of these rocks is less than 14.0 kg .
2. Explain whether it was necessary to assume that the population of the masses of these rocks is normally distributed.
3. A geologist suspects that rocks in another area have a mean mass which is less than 14.2 kg . A random sample of 100 rocks in this area has sample mean 13.5 kg . Assuming that the standard deviation for rocks in this area is also 3.1 kg , test at the \(2 \%\) significance level whether the geologist is correct.

3 The length, in centimetres, of a certain type of snake is modelled by the random variable \(X\) with mean 52 and standard deviation 6.1. A random sample of 75 snakes is selected, and the sample mean, \(\bar { X }\), is found.

1. Find \(\mathrm { P } ( 51 < \bar { X } < 53 )\).
2. Explain why it was necessary to use the Central Limit theorem in the solution to part (i).

2 The mean and standard deviation of the time spent by people in a certain library are 29 minutes and 6 minutes respectively.

1. Find the probability that the mean time spent in the library by a random sample of 120 people is more than 30 minutes.
2. Explain whether it was necessary to assume that the time spent by people in the library is normally distributed in the solution to part (i).

2 Over a long period of time it is found that the amount of sunshine on any day in a particular town in Spain has mean 6.7 hours and standard deviation 3.1 hours.

1. Find the probability that the mean amount of sunshine over a random sample of 300 days is between 6.5 and 6.8 hours.
2. Give a reason why it is not necessary to assume that the daily amount of sunshine is normally distributed in order to carry out the calculation in part (i).

2 The mean and standard deviation of the time spent by people in a certain library are 29 minutes and 6 minutes respectively.

1. Find the probability that the mean time spent in the library by a random sample of 120 people is more than 30 minutes.
2. Explain whether it was necessary to assume that the time spent by people in the library is normally distributed in the solution to part (i).

6 The continuous random variable \(R\) has the distribution \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). The results of 100 observations of \(R\) are summarised by $$\Sigma r = 3360.0 , \quad \Sigma r ^ { 2 } = 115782.84 .$$

1. Calculate an unbiased estimate of \(\mu\) and an unbiased estimate of \(\sigma ^ { 2 }\).
2. The mean of 9 observations of \(R\) is denoted by \(\bar { R }\). Calculate an estimate of \(\mathrm { P } ( \bar { R } > 32.0 )\).
3. Explain whether you need to use the Central Limit Theorem in your answer to part (ii).

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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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.

1. 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.
2. 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.

5 The random variable \(X\) has a binomial distribution with parameters \(n\) and \(p\), where \(p > 0.5\). A random sample of \(4 n\) observations of \(X\) is taken and \(\bar { X }\) denotes the sample mean. It is given that \(\mathrm { E } ( \bar { X } ) = 180\) and \(\operatorname { Var } ( \bar { X } ) = 0.0225\).

1. Find
   1. the values of \(p\) and \(n\),
   2. \(\mathrm { P } ( \bar { X } < 179.8 )\),
   3. the value of \(a\) for which \(\mathrm { P } ( 180 - a < \bar { X } < 180 + a ) = 0.99\), giving your answer correct to 2 decimal places.
   4. State how you have used the Central Limit Theorem in part (i).

The time, in minutes, taken by students to complete a test has the distribution \(\text{N}(125, 36)\).

1. Find the probability that the mean time taken to complete the test by a random sample of 40 students is less than 123 minutes. [3]
2. Explain whether it was necessary to use the Central Limit theorem in the solution to part (a). [1]

A report on the health and nutrition of a population stated that the mean height of three-year old children is 90 cm and the standard deviation is 5 cm. A sample of 100 three-year old children was chosen from the population.

1. Write down the approximate distribution of the sample mean height. Give a reason for your answer. [3]
2. Hence find the probability that the sample mean height is at least 91 cm. [3]