Question

A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population mean and standard deviation of 90 lbs. and 24 lbs., respectively, then based on a sample size of 36 boxes, what is the probability that the average weight of the boxes will exceed 94 lbs.

Answer 1

Answer:

The probability that the average weight of the boxes will exceed 94 lbs is 0.1587.

Step-by-step explanation:

Let X = weight of the boxes shipped.

The mean weight is, μ = 90 lbs and the standard deviation, σ = 24 lbs.

A sample of n = 36 boxes is selected.

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

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

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

And the standard deviation of the sample means (also known as the standard error) is given by,

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

Hence the sampling distribution of the sample mean weight of boxes is Normally distributed.

Compute the probability that the average weight of the boxes will exceed 94 lbs as follows:

$P(\bar X>94)=P(\frac{\bar X-\mu_{\bar x}}{\sigma/\sqrt{n}}>\frac{94-90}{24/\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the probability that the average weight of the boxes will exceed 94 lbs is 0.1587.