x has a normal distribution with a mean of 80.0 and a standard deviation of 3.5 find p(78.0 < x < 83.0 ) p(x>85.0) please show work

Expert Answers

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We are given a normal distribution with `bar(x)=80,s=3.5` :

(a) Find `P(78.0<x<83.0)`

First convert the data to `z` scores using `z=(x-bar(x))/s` :

`78:z=(78-80)/3.5=-2/3.5 = -0.57`


Then `P(78<x<83)=P(-.57<z<.86)`

Using a standard normal table we find the area to the left of z=-.57 to be .2843 and the area to the left of z=0.86 to be .8051

The area between them is .8051-.2843=.5208

Using a graphing utility I got .5207667191

Thus `P(78.0<x<83.0)=.521` or 52.1%

(b) `P(x>85)` First convert 85 to a z score:


Then `P(x>85)=P(z>1.43)=.0764` (If the standard normal table gives the area to the left, then 1-.9236=.0764)

Thus `P(x>85)=.076` or 7.6%

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