Leave answers as decimals (not percents). When necessary, round final answers to

Question

Leave answers as decimals (not percents). When necessary, round final answers to 4 decimal places.

1. The lifespan of a certain species of fruit fly is so short that it needs to be measured in hours. The mean lifespan is 18 hours, the standard deviation is 3.5 hours, and a graph of the frequency distribution appears to be bell shaped. A group of 25 randomly selected fruit flies is obtained. Let X be the lifespan of a fruit fly from this group in hours. a. Choose one: X is a binomial random variable X is a normal random variable b. Find P(x

Explanation / Answer

1)a) X is normal random variable

b) P(X < 20) = P((X - mean)/sd < (20 - mean)/sd)

                     = P(Z < (20 - 18)/3.5)

                     = P(Z < 0.57)

                     = 0.7157

c) P(15 < X 20)

= P((15 - 18)/3.5 < Z < (20 - 18)/3.5)

= P(-0.86 < Z < 0.57)

= P(Z < 0.57) - P(Z < -0.86)

= 0.7157 - 0.1949

= 0.5208

d) Mean of the sample mean = 18

e) SE = 3.5/sqrt(25) = 0.7

f) Option - III

                   

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