Respuesta :
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
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
[tex]n=400[/tex]
Now, calculating the probability that the average salary of the 400 players that year exceeded $1.1 million is,
[tex]P(x > 1.1)=1-P(x < 1.1\\=1-P(z < -12.5\\=1-NORMSDIST(-12.5)\\=1-0.0000\\=1.0000[/tex]
Learn more about the topic Mean:
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