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A wildlife conservationist specializing in wolf conservation read an article from 2008 that claimed that the number of wolves
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Answer #1

Given Mean \mu = 4.8 puppies

Standard Deviation \sigma = 1.2 puppies

We need to find proportion of wolf litters that have will have more than 6 puppies

P(X > 6)

z-score = (X - \mu ) / \sigma

= (6 - 4.8) / 1.2

= 1.2 / 1.2

= 1

The area to the left of z-score will give us P(X < 6) here

P(X < 6) = 0.84134 from below attached table

we know that P(X < 6) + P(X > 6) = 1

P(X > 6) = 1 - P(X < 6)

= 1 - 0.84134

= 0.15866

proportion of wolf litters that have will have more than 6 puppies = 0.15866

= 0.159 rounded to 3 decimal places   

5430 STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score. 2 .00 .01 .04 .05 .06 .07 .09 0.0

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