Question

One year, the distribution of salaries for professional sports players had mean $1.6 million and standard deviation $0.8 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players that year exceeded $1.1 million.

Answer 1

Answer:

1

Step-by-step explanation:

One year, the distribution of salaries for professional sports players had mean $1.6 million and standard deviation $0.8 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players that year exceeded $1.1 million.

z = (x-μ)/σ/√n

where

x is the raw score

μ is the population mean

σ is the population standard deviation

n = random number of samples

Mean $1.6 million and Standard deviation $0.8 million.

z = (1100000- 1600000)/800000/√400

z = -500000/800000/20

z = -500000/40000

z = -12.5

Probability value from z score table is :

P(x ≤ Z) = P(x = 1100000) = P(z = 12.5)

= 0

P(x>Z) = 1 - P(x < Z)

= 1 - 0

= 1

The approximate probability that the average salary of the 400 players that year exceeded $1.1 million is 1

Answer 2

Hence, the required probability is 1.

Mean:

The mean is one of the measures of central tendency.

Given mean is 1.6 and the standard deviation is 0.8

$n=400$

Now, calculating the probability that the average salary of the 400 players that year exceeded $1.1 million is,

$P(x > 1.1)=1-P(x < 1.1\\=1-P(z < -12.5\\=1-NORMSDIST(-12.5)\\=1-0.0000\\=1.0000$

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