Major league baseball salaries averaged $3.26 million with a standard deviation

Question

Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million in a recent year. Suppose a sample of 100 major league players was taken. Find the approximate probability that the mean salary of the 100 players was no more than $3.5 million. approximately 0 0.0228 0.9772 approximately 1 Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million in a recent year. Suppose a sample of 100 major league players was taken. Find the approximate probability that the mean salary of the 100 players was no more than $4.0 million. approximately 0 0.0228 0.9772 approximately 1 Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million in a recent year. Suppose a sample of 100 major league players was taken. Find the approximate probability that the mean salary of the 100 players was no more than $3.0 million. approximately 0 0.0151 0.9849 approximately 1 At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean? 0.029 0.050 0.091 0.120

Explanation / Answer

Q1.
Mean ( u ) =3.26
Standard Deviation ( sd )= 1.2/ Sqrt(n) = 0.12
Number ( n ) = 100
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X > 3.5) = (3.5-3.26)/1.2/ Sqrt ( 100 )
= 0.24/0.12= 2
= P ( Z >2) From Standard Normal Table
= 0.0228                  
Q2.
P(X > 4) = (4-3.26)/1.2/ Sqrt ( 100 )
= 0.74/0.12= 6.1667
= P ( Z >6.1667) From Standard Normal Table
= 0                  
Q3.
P(X > 3) = (3-3.26)/1.2/ Sqrt ( 100 )
= -0.26/0.12= -2.1667
= P ( Z >-2.1667) From Standard Normal Table
= 0.9849                  
P(X < = 3) = (1 - P(X > 3)
= 1 - 0.9849 = 0.0151

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