Question

A large car dealership claims that 47% of its customers are looking to buy a sport utility vehicle (SUV). A random sample of 61 customers is surveyed. What is the probability that less than 40% of the sample are looking to buy an SUV?

Answer 1

Using the normal distribution and the central limit theorem, it is found that the probability is of 0.1368 = 13.68%.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

In this problem:

• 47% of its customers are looking to buy a sport utility vehicle (SUV), hence p = 0.47.

• A sample of 61 customers is taken, hence n = 61.

The mean and the standard error are given by:

$\mu = p = 0.47$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.47(0.53)}{61}} = 0.0639$

The probability that less than 40% of the sample are looking to buy an SUV is the p-value of Z when X = 0.4, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.4 - 0.47}{0.0639}$

Z = -1.095

Z = -1.095 has a p-value of 0.1368.

0.1368 = 13.68% probability that less than 40% of the sample are looking to buy an SUV.

To learn more about the normal distribution and the central limit theorem, you can check brainly.com/question/24663213