Question

the diameter of small nerf balls manufactured overseas 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

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

Normal probability distribution

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

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

Z = X-u/σ

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  and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean u  and standard deviation s=σ/$\sqrt{n}$

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:

u= 5.2 ,σ = 0.08 , n= 20 s= 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= X-u/σ

By the central Limit Theorem

Z= X-u/s

X= 5.2358

Z = -2

then

X= 5.1642

learn more about of interval here

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