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) =
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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