Question

In: Statistics and Probability

Assume that women's weights are normally distributed with a mean given by μ=143 lb and a...

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

Solutions

Expert Solution

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


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