Question

The baseball players salaries from the 2015 season have a population mean μ = 4.215 million dollars and the population standard deviation σ = 5.481 million dollars.

1. Use the equation for central limit theorem to determine the mean and the standard deviation of the sampling distribution of sample means of n = 60 randomly selected baseball players

2. The Central Limit Theorem indicates that the sampling distribution is approximately normal. Why does this allow us to use the NORM.DIST and NORM.INV functions?

3. Use the NORM.DIST function together with the values in part a to find the probability that a randomly selected sample of 60 baseball players will have a sample mean salary between 3 million and 5 million dollars?

4. Use the NORM.DIST function together with the values in part a to find the probability that a randomly selected sample of 60 baseball players will have a sample mean salary greater than 6.5 million dollars?

5. Use the NORM.INV function together with the values in part a to find the value xR for which 25% of randomly selected samples of 50 baseball players will have a mean salary smaller than xR?

6. Use the previous part to determine the value xL for which 25% of randomly selected samples of 50 baseball players will have a mean salary greater than xL?

Answer 1

The baseball players salaries from the 2015 season have a population mean μ = 4.215 million dollars and the population standard deviation σ = 5.481 million dollars.

let 'X' represent the salaries of the players

X N( =4.215, =5.481   )

z-score =   

Use the equation for central limit theorem to determine the mean and the standard deviation of the sampling distribution of sample means of n = 60 randomly selected baseball players

Central limit theorem states that if the sample size is large ( n > 30) then the distribution of the means of similar sample size will approximately follow normal distribution. The parameters are as follows.

n = 60

z-score= ( - 4.215) / 0.7076

The Central Limit Theorem indicates that the sampling distribution is approximately normal. Why does this allow us to use the NORM.DIST and NORM.INV functions?

Central limit theorem states that if the sample size is large ( n > 30) then the distribution of the means of similar sample size will approximately follow normal distribution.

Use the NORM.DIST function together with the values in part a to find the probability that a randomly selected sample of 60 baseball players will have a sample mean salary between 3 million and 5 million dollars?

P(3 < < 5) = P( < 5 ) - P( < 3)

So these are two cdf. To use the norm.dist we input : norm.dist(x,mean, sd, true)

where the last argument is for finding the cumulative probability.

Therefore here we sub

P(3 < < 5) = P( < 5 ) - P( < 3)

= norm.dist(5, 4.215, 0.7076 true) - norm.dist(3, 4.215, 0.7076 true)

= 0.86637 - 0.04298

Ans:0.82339

Use the NORM.DIST function together with the values in part a to find the probability that a randomly selected sample of 60 baseball players will have a sample mean salary greater than 6.5 million dollars?

P( > 6.5) = 1 - P( < 6.5)

=1 - norm.dist(6.5, 4.215, 0.7076 true)

=1 - 0.99938

ans: 0.00621

Use the NORM.INV function together with the values in part a to find the value xR for which 25% of randomly selected samples of 50 baseball players will have a mean salary smaller than xR?

THe distribution will be changed since now we have n = 50

Here we want to find the mean or the 'x' value from the given probability.

P( X < ) = 25%

For this we use: norm.dist(p.,mean, sd)

P( X < ) = 25%

=norm.dist(0.25.,4.215, 0.77513)

ans: 3.69218

Use the previous part to determine the value xL for which 25% of randomly selected samples of 50 baseball players will have a mean salary greater than xL?

P( X > ) = 25%

For this we use: norm.dist(p.,mean, sd)

P( X < ) = 75% ...............we subtract from 1 so we get '<' sign since values are for less than type.

=norm.dist(0.75.,4.215, 0.77513)

ans: 4.73782