Question

The diameter of small nerf balls manufactured at a factory in china is expected to be approximately normally distributed with a mean of 5.2 inches and a standard deviation of .08 inches. suppose a random sample of 20 balls is selected. find the interval that contains 95.44 percent of the sample means.

Answer 1

That bank must have a lot of moneu but im still looking for the answer

Answer 2

Answer:

The interval that contains 95.44 percent of the sample means is between 5.1642 inches and 5.2358 inches

Step-by-step explanation:

We need to understand the normal probability distribution and the central limit theorem to solve this question.

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:

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

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 5.2, \sigma = 0.08, n = 20, s = \frac{0.08}{\sqrt{20}} = 0.0179$

Find the interval that contains 95.44 percent of the sample means.

0.5 - (0.9544/2) = 0.0228

Pvalue of 0.0228 when Z = -2.

0.5 + (0.9544/2) = 0.9772

Pvalue of 0.9772 when Z = 2.

So the interval is from X when Z = -2 to X when Z = 2

Z = 2

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

By the Central Limit Theorem

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

$2 = \frac{X - 5.2}{0.0179}$

$X - 5.2 = 2*0.0179$

$X = 5.2358$

Z = -2

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

$-2 = \frac{X - 5.2}{0.0179}$

$X - 5.2 = -2*0.0179$

$X = 5.1642$

The interval that contains 95.44 percent of the sample means is between 5.1642 inches and 5.2358 inches