Question 655186
1.)Find the area under the standard normal curve: 
I. to the right of z = -2.32
Using a TI-84 I get: normalcdf(-2.32,100) = 0.9898
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II. to the left of z = -2.32
normalcdf(-100,-2.32) = 0.0102
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2.) Assume that the population of heights of male college students is approximately normally distributed with mean m of 70 inches and standard deviation s of 3.75 inches. Show all work. 
(A) Find the proportion of male college students whose height is greater than 68 inches.
z(68) = (68-70)/3.75 = -0.5333
P(x > 68) = P(z > -0.5333) = 0.7031
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2) Find the proportion of male students whose height is less than 68 
Ans: 1 - 0.7031 = 0.2969
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3.) The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.5 inches and a standard deviation of 0.50 inches. Show all work. 
(A) What percentage of the grapefruits in this orchard have diameters less than 6.3 inches?
z(6.3) = (6.3-6.5)/0.5 = -0.4
P(x < 6.3) = P(z < -0.4) = normalcdf(-100,-0.4) = 0.3446
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(B) What percentage of the grapefruits in this orchard are larger than 6.15 inches?
z(6.15) = (6.15-6.5)/0.5 = 0.7
P(x > 6.15) = P(z > 0.7) = 0.2420
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Cheers,
Stan H.
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