Question

1. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 116.7-cm and a standard deviation of 1.8-cm. For shipment, 22 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is less than 116.8-cm.
P( x < 116.8-cm) =

2. CNNBC recently reported that the mean annual cost of auto insurance is 1010 dollars. Assume the standard deviation is 216 dollars. You take a simple random sample of 84 auto insurance policies.

Find the probability that a single randomly selected value is less than 974 dollars.
P(X < 974) =

Find the probability that a sample of size n=84n=84 is randomly selected with a mean less than 974 dollars.
P( x < 974) =

Answer 1

This is a normal distribution question with

z = 0.2606

This implies that

P(x < 116.8) = P(z < 0.2606) = 0.6028

2)

a)

a P(x < 974.0)=?

The z-score at x = 974.0 is,

z = 974.0-1010.0/216.0

z = -0.1667

This implies that

P(x < 974.0) = P(z < -0.1667) = 0.4338

b)

This is a normal distribution question with

The z-score at x = 974.0 is,

z = -1.5275

This implies that

P(x < 974.0) = P(z < -1.5275) = 0.0633

PS: you have to refer z score table to find the final probabilities.

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