In: Statistics and Probability
Assume that women's weights are normally distributed with a mean
given by μ=143 lb and a standard deviation given by σ=29 lb.
(a) If 1 woman is randomly selected, find the
probability that her weight is above 177
(b) If 6 women are randomly selected, find the
probability that they have a mean weight above 177
(c) If 78 women are randomly selected, find the
probability that they have a mean weight above 177
Solution :
Given that ,
(a)
P(x > 177) = 1 - P(x < 177)
= 1 - P[(x - ) / < (177 - 143) / 29)
= 1 - P(z < 1.31)
= 1 - 0.9049
= 0.0951
Probability = 0.0951
(b)
= / n = 29 / 6 = 11.8392
P( > 177) = 1 - P( < 177)
= 1 - P[( - ) / < (177 - 143) / 11.8392]
= 1 - P(z < 2.87)
= 1 - 0.9979
= 0.0021
Probability = 0.0021
(c)
= / n = 29 / 78 = 3.2836
P( > 177) = 1 - P( < 177)
= 1 - P[( - ) / < (177 - 143) / 3.2836]
= 1 - P(z < 10.35)
= 1 - 1
= 0
Probability = 0