Question

1. A distribution of values is normal with a mean of 110.8 and a
standard deviation of 33.5.

Find the probability that a randomly selected value is less than
20.7.

*P*(*X* < 20.7) =

Enter your answer as a number accurate to 4 decimal places. *Note:
all z-scores must be rounded to the nearest hundredth.

2. A distribution of values is normal with a mean of 2368.9 and
a standard deviation of 39.4.

Find the probability that a randomly selected value is greater than
2259.8.

*P*(*X* > 2259.8) =

Enter your answer as a number accurate to 4 decimal places. *Note:
all z-scores must be rounded to the nearest hundredth.

3. A distribution of values is normal with a mean of 230.4 and a
standard deviation of 27.8.

Find the probability that a randomly selected value is between
177.9 and 279.1.

*P*(177.9 < *X* < 279.1) =

Enter your answer as a number accurate to 4 decimal places. *Note:
all z-scores must be rounded to the nearest hundredth.

4. The number of miles a certain type of brake pad will last is
normally distributed with a mean of 50000 miles and standard
deviation 2020 miles. Find the probability that, if you install
this type of brake pad, it will last more than 53575 miles.

*P*(*X* > 53575) =

Enter your answer as a number accurate to 4 decimal places. *Note:
all z-scores must be rounded to the nearest hundredth.

5.

The heights of women aged 20-29 are normally distributed with
mean 64.2 inches and standard deviation 3.8 inches. What
**percent** of women are below 72.6 inches? Round to
the nearest hundredth of a percent.

Enter your answer as a **percent** rounded to the
nearest hundredth. Do not enter the percent symbol.

*Note: all z-scores must be rounded to the nearest hundredth.

Answer #1

1.

P(X < 20.7) = P[Z < (20.7 - 110.8) / 33.5]

= P[Z < -2.69]

= 0.0036

2.

P(X > 2259.8) = P[Z > (2259.8 - 2368.9)/39.4]

= P[Z > -2.77]

= 0.9972

3.

P(177.9 < X < 279.1) = P(X < 279.1) - P(X <
177.9)

= P[Z < (279.1 - 230.4) / 27.8] - P[Z < (177.9 - 230.4) /
27.8]

= P[Z < 1.75] - P[Z < -1.89]

= 0.9599 - 0.0294

= 0.9305

4.

P(X > 53575) = P[Z > (53575 - 50000)/2020]

= P[Z > 1.77]

= 0.0384

5.

P(X < 72.6) = P[Z < (72.6 - 64.2) / 3.8]

= P[Z < 2.21]

= 0.9864

= 98.64 %

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