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sdas [7]
2 years ago
12

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 ?​
Mathematics
1 answer:
xxTIMURxx [149]2 years ago
6 0

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

<h3>Normal Probability Distribution</h3>

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 <u>one subtracted by the p-value of Z when X = 80</u>, 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 brainly.com/question/4079902

#SPJ1

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