# Normal probability distributions, proportions, percentages

1. A set of 50 data values has a mean of 44 and a variance of 16.

I. Find the standard score (z) for a data value = 51.

II. Find the probability of a data value < 51.

2. Find the area under the standard normal curve:

I. to the right of z = -1.34

II. to the left of z = -1.34

3. Assume that the population of heights of female college students is approximately normally distributed with mean  of 67 inches and standard deviation  of 3.95 inches. Show all work.

(A) Find the proportion of female college students whose height is greater than 63 inches.

(B) Find the proportion of female college students whose height is no more than 63 inches.

4. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.35 inches and a standard deviation of 0.55 inches. Show all work.

(A) What percentage of the grapefruits in this orchard have diameters less than 6.7 inches?

(B) What percentage of the grapefruits in this orchard are larger than 6.85 inches?

5. Find the normal approximation for the binomial probability that x = 4, where n = 12 and p = 0.7. Compare this probability to the value of P(x=4) found in Table 2 of Appendix B in your textbook

6. A set of data is normally distributed with a mean of 1000 and standard deviation of 100.

?What would be the standard score for a score of 800?

?What percentage of scores is between 1000 and 800?

?What would be the percentile rank for a score of 800?

#### Solution Preview

1. A set of 50 data values has a mean of 44 and a variance of 16.

I. Find the standard score (z) for a data value = 51.

II. Find the probability of a data value < 51.

(I) z = (x - μ)/s/√n) = (51 - 44)/(4/√50) = 12.37

(II) P(x < 51) = P(z < 12.37) = 1

2. Find the area under the standard normal curve:

I. to the right of z = -1.34

II. to the left of z = -1.34

(I) 0.91

(II) 0.09

3. Assume that the population of heights of female college students is approximately normally distributed with mean of 67 inches and standard deviation of 3.95 inches. Show all work.

(A) ...

#### Solution Summary

Complete, Neat and Step-by-step Solutions are provided in the attached file. Normal probability distributions, proportions and percentages are determined for different data sets.