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1. A set of 50 data values has a mean of 28 and a variance of 4.
a. Find the standard score (z) for a data value = 31.
z=
31− 28 3
= = 1.5
2
4
b. Find the probability of a data value > 31.
P(x > 31) = P(z > 1.5) = 0.0668
2. Find the area under the standard normal curve:
a. to the right of z = 2.37
0.0089
(This value comes from an online calculator.)
b. to the left of z = 2.37
0.9911
(This value comes from an online calculator. No other work to show)
Note that the answers to part a and part b add up to 1.00, as they should.
3. Assume that the population of heights of female college students is approximately
normally distributed with mean m of 67 inches and standard deviation s of 3.95 inches.
Show all work.
(A) Find the proportion of female college students whose height is greater than 63 inches.
z=
63− 67
= −1.01266
3.95
P(x > 63) = P(z > -1.01266) = 0.8444
(B) Find the proportion of female college students whose height is no more than 63
inches.
P(x ≤ 63) = 1 – P(x > 63) = 1 – 0.8444 = 0.1556
4. The diameters of grapefruits in a certain orchard are normally distributed with a mean
of 6.45 inches and a standard deviation of 0.45 inches. Show all work.
(A) What percentage of the grapefruits in this orchard have diameters less than 7.1
inches?
z=
7.1− 6.45
= 1.4444
0.45
P(x < 7.1) = P(z < 1.4444) = 0.9257
Convert this proportion to a percentage by multiplying by 100%:
0.9257 * 100% = 92.57%
(B) What percentage of the grapefruits in this orchard are larger than 6.8 inches?
z=
6.8 − 6.45
= 0.7778
0.45
P(x > 6.8) = P(z > 0.7778) = 0.2183
Convert this proportion to a percentage by multiplying by 100%:
0.2183 * 100% = 21.83%
5. Find the normal approximation for the binomial probability that x = 4, where n = 13
and p = 0.3. Compare this probability to the value of P(x=5) found in Table 2 of
Appendix B in your textbook.
µ = np = (13) ( 0.3) = 3.9
σ = npq =
(13) (0.3) (0.7) = 1.6523
z1 =
( 4.5 − 3.9) = 0.3613
z1 =
(3.5 − 3.9) = −0.2421
1.6523
1.6523
P(x = 4) ≈ P(3.5 < x < 4.5) = P(-0.2421 < z < 0.3613)
= P(z < 0.3613) – P(z < -0.2421)
= 0.6411 - 0.4044
= 0.2367
The value from the table for P(x = 5) is 0.180
The value from the table for P(x = 4) is 0.234
6. A set of data is normally distributed with a mean of 100 and standard deviation of 15.
· What would be the standard score for a score of 70?
z=
70 −100
= −2
15
· What percentage of scores is between 100 and 70?
P(70 < x < 100) = P(-2 < z < 0) = P(z < 0) – P(z < -2) = 0.5 – 0.0228 = 0.4772
Convert this to a percentage by multiplying by 100%:
0.4772 * 100% = 47.72%
· What would be the percentile rank for a score of 70?
The percentage of scores below 70 is:
P(x < 70) = P(z < -2) = 0.0228
0.0288 * 100% = 2.88% ≈ 3rd percentile.