Mathematics, 11.06.2021 18:40 warnene17

The distribution of weights of adult males in a certain county is strongly right-skewed with a mean weight of 185 lbs and standard deviation 16 lbs. What is the probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs

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0.962 = 96.2% probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

The distribution of weights of adult males in a certain county is strongly right-skewed with a mean weight of 185 lbs and standard deviation 16 lbs.

This means that $\mu = 185, \sigma = 16$

Sample of 100:

This means that $n = 100, s = \frac{16}{\sqrt{100}} = 1.6$

What is the probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs?

This is the p-value of Z when X = 188 subtracted by the p-value of Z when X = 172. So

X = 188

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

By the Central Limit Theorem

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

$Z = \frac{188 - 185}{1.6}$

Z = 1.875

Z = 1.875 has a p-value of 0.9620

X = 172

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

$Z = \frac{172 - 185}{1.6}$

Z = -8.125

Z = -8.125 has a p-value of 0.

0.9620 - 0 = 0.962

0.962 = 96.2% probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs.

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