Question

Suppose a simple random sample of size n = 36 is obtained from a population that is skewed right with μ = 75 and o = 24. What is P x > 80 ? Suppose a simple random sample of size n = 36 is obtained from a population that is skewed right with μ = 75 and o = 24. What is P x > 80 ?​

Answer 1

Using the normal distribution, it is found that the probability that the sample mean is above 80 is of 0.1056 = 10.56%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

The parameters are given as follows:

$\mu = 75, \sigma = 24, n = 36, s = \frac{24}{\sqrt{36}} = 4$.

The probability that the sample mean is above 80 is one subtracted by the p-value of Z when X = 80, hence:

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

By the Central Limit Theorem

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

$Z = \frac{80 - 75}{4}$

Z = 1.25

Z = 1.25 has a p-value of 0.8944.

1 - 0.8944 = 0.1056.

0.1056 = 10.56% probability that the sample mean is greater than 80.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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