The breaking strength of a rivet has a mean value of 10,050 psi and a standard deviation of 499 psi. (a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9,950 and 10,250? (Round your answer to four decimal places.)

P(9950 ≤ x ≤ 10250) = 0.54

Step-by-step explanation:

We would assume a normal distribution for the breaking strength of a rivet. We would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

n = number of samples

x = Breaking strengths of rivet.

µ = mean breaking strength

σ = standard deviation

From the information given,

µ = 10,050 psi

σ = 499 psi

n = 40

The probability that the sample mean breaking strength for a random sample of 40 rivets is between 9,950 and 10,250 is expressed as

P(9950 ≤ x ≤ 10250)

For x = 9950,

z = (9950 - 10050)/499/√40 = - 0.13

Looking at the normal distribution table, the probability corresponding to the z score is 0.45

For x = 10250,

z = (10250 - 10050)/499/√40 = 2.54

Looking at the normal distribution table, the probability corresponding to the z score is 0.99

Therefore,

P(9950 ≤ x ≤ 10250) = 0.99 - 0.45 = 0.54

30005

yes ok