IQ scores in a certain population are normally distributed with a mean of 98 and
ID: 3225914 • Letter: I
Question
IQ scores in a certain population are normally distributed with a mean of 98 and a standard deviation of 19. (Give your answers correct to four decimal places.) (a) Find the probability that a randomly selected person will have an IQ score between 87 and 99.
(b) Find the probability that a randomly selected person will have an IQ score above 97.
The 70-year long-term record for weather shows that for New York State, the annual precipitation has a mean of 39.03 inches and a standard deviation of 4.47 inches. The annual precipitation amount has a normal distribution. (Give your answers correct to four decimal places.)
(b) What is the probability that next year the total precipitation for New York State is between 43 and 48.4 inches?
Find the area under the standard normal curve between z = 0.74 and z = 2.3, P(0.74 < z < 2.3). (Give your answer correct to four decimal places.)
Consider the following. (Give your answers correct to four decimal places.)
(b) Find P(-1.76 < z < 2.03).
Explanation / Answer
Mean = 98
Standard deviation = 19
a) P(87<x<99) = P(X<99) - P(X<87)
= P(Z < (99-98)/19) - P(Z < (87-98)/19)
= P(Z < 0.05) - P(Z<-0.58)
= 0.5199 - 0.2810
= 0.2389
b) P(X>97) = 1 - P(X<97)
= 1 - P(Z < (97-98)/19)
= 1 - P(Z < -0.050
= 1 - 0.4801
= 0.5199
Mean = 39.03 inches
Standard deviation = 4.47 inches
b) P(43<X<48.4) = P(X<48.40 - P(X<43)
= P(Z < (48.4-39.03)/4.47) - P(Z < (43-39.03)/4.47)
= P(Z < 2.10) - P(Z<0.89)
= 0.9821 - 0.8133
= 0.1688
P (0.74 < z < 2.3) = P(z < 2.3) - P(z< 0.74)
= 0.9893 - 0.7704
= 0.2189
(b) P(-1.76 < z < 2.03) = P(z < 2.3) - P(z< -1.76)
= 0.9893 - 0.0392
= 0.9501
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