Question

If the mean GPA among students is 3.25 with a standard deviation of 0.75, what is the probability that a random sample of 300 students will have a mean GPA greater than 3.30

Answer 1

The value is
`P(X > 3.30) = 0.12405`

Explanation:

From the question we are told that

The mean GPA is
`\mu = 3.25`

The standard deviation is
`\sigma = 0.75`

The sample size is n = 300

Generally the standard error of mean is mathematically represented as

`\sigma_(\= x) = (\sigma )/(√(n) )`

=>
`\sigma_(\= x) = (0.75)/(√(300) )`

=>
`\sigma_(\= x) = 0.0433`

Generally the probability that a random sample of 300 students will have a mean GPA greater than 3.30 is mathematically represented as

`P(X > 3.30) = P((X - \mu)/(\sigma_(\= x)) > (3.30 -3.25)/( 0.0433) )`

`(\= X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ \= X )`

`P(X > 3.30) = P(Z> 1.155 )`

From the z table the probability of (Z > 1.155 ) is

`P(Z> 1.155 ) = 0.12405`

`P(X > 3.30) = 0.12405`