Question

2. Suppose that car's arrive at Burger King's drive-through at the rate of 20 cars every hour between 12:00 noon and 1:00 pm. A random sample of 40 one-hour time periods between 12:00 noon and 1:00pm is selected and has 22.1 as the mean number of cars arriving. a) Why is the sampling distribution of x approximately normal? b) What is the mean and standard deviation of the sampling distribution of x assuming that y = 20 and o = 720? c) What is the probability that a simple random sample of 40 one-hour time periods results in a mean of at least 22.1 cars?

Answer 1

Given =

Average rate at car arrives = 20 = Population mean

Population standard deviation = sqrt(20) = 4.472

Random sample taken of size (n) = 40

Sample mean = 22.1

a)

sampling distribution of mean would be approximately normal because of central limit theorem.

Central Limit Theorem

The central limit theorem states that:

Given a population with a finite mean μ and a finite non-zero variance σ², the sampling distribution of the mean approaches a normal distribution with a mean of μ and a variance of σ²/N as N, the sample size, increases.

b)

mean of sampling distribution is same as population mean hence

$\bar x = population\ mean = 20$

Standard deviation[ sampling distribution] = Population standard deviation / sqrt[sample size]

= 4.472 / sqrt(40)

= 0.707

c)

μ=20, σ=4.472, n=40

We need to compute Pr(Xˉ≥22.1).

The corresponding z-value needed to be computed is:

$Z = \frac{\bar X-\mu}{\sigma/\sqrt{n}} = \frac{ 22.1-20}{ 4.472/\sqrt{ 40}} = 2.9699$

Therefore, we get that

$\Pr(\bar X \geq 22.1) = \Pr\left(Z \ge \frac{ 22.1-20}{ 4.472/\sqrt{ 40}}\right) = \Pr(Z \ge 2.9699)$

$=1-0.9985=0.0015$

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