# the diameter of grapefruits in a certian orchard are normally distributed w/ a mean of 6.45 in & standard d of 0.45 in what % of the grapefruit in the orchard have diameters of less than 7.1 in what % of the grapefruit in the orhard are larger than 6.8 in please show each step and explain

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We are given a normally distributed set with `bar(x)=6.45` inches and `s=.45` inches.

(a) What percentage have diameters less than 7.1in?

Convert the raw data to a `z` score:

`z=(x-bar(x))/s=(7.1-6.45)/(.45)=1.44`

Then `P(x<7.1)=P(z<1.44)` From a standard normal table we find the percentage of values to the left of `z=1.44` to be .9251

Thus 92.5% will have diameters less than 7.1 inches.

(b) What percentage are larger than 6.8"?

Convert 6.8 to a `z` score:

`z=(6.8-6.45)/.45=.7778`

Then `P(x>6.8)=P(z>.7778)=.2183` (You can find the percentage less than .78 and subtract from 1)

The percentage with diameters greater than 6.8" is 21.8%

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