Question

A particular automobile costs an average of $21,755 in the Pacific Northwest. The standard deviation of prices is $650. Suppose a random sample of 30 dealerships in Washington and Oregon is taken, and their managers are asked what they charge for this automobile. What is the probability of getting a sample average cost of less than $21,500

Answer 1

`P(\bar X <21500)`

And we can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 21500 we got:

`z = (21500-21755)/((650)/(√(30)))= -2.149`

And we can use the normal standard distribution or excel and we got:

`P(Z<-2.149) =0.0158`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the automobile costs of a population, and for this case we know the following info

Where
`\mu=21755` and
`\sigma=650`

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And for this case we want to find this probability:]

`P(\bar X <21500)`

And we can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 21500 we got:

`z = (21500-21755)/((650)/(√(30)))= -2.149`

And we can use the normal standard distribution or excel and we got:

`P(Z<-2.149) =0.0158`

Answer 2

1.58% probability of getting a sample average cost of less than $21,500

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30

In this problem, we have that:

`\mu = 21755, \sigma = 650, n = 30, s = (650)/(√(30)) = 118.67`

What is the probability of getting a sample average cost of less than $21,500

This is 1 subtracted by the pvalue of Z when X = 21500. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (21500 - 21755)/(118.67)`

`Z = -2.15`

`Z = -2.15` has a pvalue of 0.0158

1.58% probability of getting a sample average cost of less than $21,500