Question

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. The distribution of this particular variable is very right skewed.(a) Suppose we let (X-bar) be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that (X-bar) is less than 2?(b) Now let X be the mean number of accidents per week at the intersection during a year (52 weeks). What is the probability that X is less than 2?

Answer 1

Answer:

The probability that sample mean is less than 2  is 0.1423.

Step-by-step explanation:

Let X denote the number of accidents per week at a hazardous intersection.

It is provided that the mean and standard deviation of X are, μ = 2.2 and σ = 1.4.

(a)

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.  

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

As the sample size is not large enough, i.e. n = 9 < 30, the Central Limit Theorem cannot be applied to approximate the sampling distribution of the mean number of accidents per week at the intersection.

And since the distribution of X is not specified, the probability cannot be computed.

(b)

In this case, the sample size is large enough, i.e. n = 52 > 30, the Central Limit Theorem can be applied to approximate the sampling distribution of the mean number of accidents per week at the intersection.

Compute the probability that sample mean is less than 2 as follows:

$P(\bar X$

Thus, the probability that sample mean is less than 2  is 0.1423.