The average salary for a certain profession is exist90,000. Assume that the stan

Question

The average salary for a certain profession is exist90,000. Assume that the standard deviation of such salaries is exist38,000. Consider a random 65 people in this profession and let x represent the mean salary for the sample. A. The shape is that of a uniform distribution. B. The shape is that of a poisson distribution. C. The shape is that of a binomial distribution. D. The shape is that of a normal distribution. d. Find the z-score for the value x = 82,000. (Round to two decimal places as needed.) e. Find P(x > 82,000). P(x > 82.000) = (Round to three decimal places as needed)

Explanation / Answer

Solution:-

z = (x - mu) / s/sqrt(n)

= (82,000 - 90,000) / 38,000/sqrt(65)

= -8,000 / 4713.31991439

= -1.697 or -1.70

P(X > 82000) = ?

Cumulative probability: P(X < 82000) = 0.41663

Therefore, P(X > 82000) = 1 - 0.41663 = 0.58337 or 0.583

Cumulative probability: P(X < 82000) = 0.41663

Therefore, P(X > 82000) = 1 - 0.41663 = 0.58337 or 0.583

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