Question

A population of values has a normal distribution with μ=134.6μ=134.6 and σ=75.5σ=75.5. You intend to draw a random sample of size n=225n=225.
Find the probability that a sample of size n=225n=225 is randomly selected with a mean between 122 and 129.6.
P(122 < M < 129.6) =
Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 242.6-cm and a standard deviation of 0.8-cm. For shipment, 6 steel rods are bundled together.
Find the probability that the average length of a randomly selected bundle of steel rods is greater than 243.1-cm.
P(M > 243.1-cm) =
Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

a) A population of values has a normal distribution with mean μ=134.6 and standard deviation σ=75.5. Now the probability that a sample of size n=225 is randomly selected with a mean between 122 and 129.6.
P(122 < < 129.6) is computed using the Z-score calculated as:

Thus the probability is computed as:

P(−2.503≤ Z ≤ −0.993) Now the probability is computed using the excel formula for Normal distribution which is =NORM.S.DIST(-0.993, TRUE)- NORM.S.DIST(-2.503, TRUE), thus the probability is computed as:

P(122 < < 129.6) =  0.1542

b) Given a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of = 242.6 cm and a standard deviation of σ = 0.8cm. For shipment, 6 steel rods are bundled together.
Find the probability that the average length of a randomly selected bundle of steel rods is greater than > 243.1 cm.
P( > 243.1cm) is calculated by finding the Z score at Mean = 243.1 as:

Thus the probability is computed as P( Z> 1.531) which is computed using the excel formula for normal distribution which is =1-NORM.S.DIST(1.531, TRUE), thus the probability is computed as 0.0629.