Question

1) A company produces steel rods. The lengths of the steel rods
are normally distributed with a mean of 120.4-cm and a standard
deviation of 1.4-cm.

Find the probability that the length of a randomly selected steel
rod is less than 120.1-cm.

*P*(*X* < 120.1-cm) =

2) A company produces steel rods. The lengths of the steel rods
are normally distributed with a mean of 236.7-cm and a standard
deviation of 2.4-cm. For shipment, 10 steel rods are bundled
together.

Find *P*_{18}, which is the average length
separating the smallest 18% bundles from the largest 82%
bundles.

*P*_{18} =

3) A population of values has a normal distribution with
μ=132.5μ=132.5 and σ=20.6σ=20.6. You intend to draw a random sample
of size n=119n=119.

Find *P*_{38}, which is the mean separating the
bottom 38% means from the top 62% means.

*P*_{38} (for sample means) =

4) A population of values has a normal distribution with
μ=109.1μ=109.1 and σ=34.1σ=34.1. You intend to draw a random sample
of size n=208n=208.

Find *P*_{7}, which is the score separating the
bottom 7% scores from the top 93% scores.

*P*_{7} (for single values) =

Find *P*_{7}, which is the mean separating the
bottom 7% means from the top 93% means.

*P*_{7} (for sample means) =

5) CNNBC recently reported that the mean annual cost of auto
insurance is 961 dollars. Assume the standard deviation is 294
dollars. You take a simple random sample of 67 auto insurance
policies.

Find the probability that a single randomly selected value is at
least 975 dollars.

*P*(*X* > 975) =

Find the probability that a sample of size n=67n=67 is randomly
selected with a mean that is at least 975 dollars.

*P*(*M* > 975) =

Answer #1

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....

A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 207.9-cm and a standard
deviation of 1.5-cm. For shipment, 26 steel rods are bundled
together.
Find P26,
which is the average length separating the smallest 26% bundles
from the largest 74% bundles.

A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 225.1-cm and a standard
deviation of 1.3-cm. For shipment, 10 steel rods are bundled
together. Find P95, which is the average length separating the
smallest 95% bundles from the largest 5% bundles. P95 = _____-cm
Enter your answer as a number accurate to 2 decimal place.

A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 197.5-cm and a standard
deviation of 2-cm. For shipment, 6 steel rods are bundled
together.
Find P11, which is the average length separating the smallest
11% bundles from the largest 89% bundles.
P11 =______________ -cm
Enter your answer as a number accurate to 2 decimal place.
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 179.5-cm and a standard
deviation of 0.7-cm. For shipment, 20 steel rods are bundled
together.
Find P95, which is the average length
separating the smallest 95% bundles from the largest 5%
bundles.
P95 = -cm
Enter your answer as a number accurate to 2 decimal place. 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 216.5-cm and a standard
deviation of 2.3-cm. For shipment, 24 steel rods are bundled
together.
Find P15, which is the average length
separating the smallest 15% bundles from the largest 85%
bundles.
P15 = -cm
Enter your answer as a number accurate to 2 decimal place. 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 252.7-cm and a standard
deviation of 2.4-cm. For shipment, 9 steel rods are bundled
together.
Find P82, which is the average length
separating the smallest 82% bundles from the largest 18%
bundles.
P82 = -cm
Enter your answer as a number accurate to 2 decimal place. 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 98.8 cm and a standard
deviation of 2.5 cm. For shipment, 22 steel rods are bundled
together.
Note: Even though our sample size is less than 30, we can use
the z score because
1) The population is normally distributed and
2) We know the population standard deviation, sigma.
Find the probability that the average length of a randomly selected
bundle of...

A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 91.1-cm and a standard
deviation of 0.5-cm. For shipment, 25 steel rods are bundled
together.
Find the probability that the average length of a randomly selected
bundle of steel rods is greater than 90.8-cm.
P(M > 90.8-cm) =

A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 226.6-cm and a standard
deviation of 1.7-cm. For shipment, 10 steel rods are bundled
together. Find the probability that the average length of a
randomly selected bundle of steel rods is less than 227.9-cm. P(M
< 227.9-cm) =

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