Mathematics, 30.07.2019 07:00 Zachgrainger4436

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.

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Answer from: izquierdohannah

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

Answer from: Quest

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

Answer from: Quest

In this problem it decreased 20%

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