gradebelow the 60th percentile of this distr CUIES, and an F grade the 60th perc

Question

gradebelow the 60th percentile of this distr CUIES, and an F grade the 60th percentile of this distribu- o all sCoeach possible letter grade, find the lowest to all tion. s table score within the established ran accepthe lowest acceptable score for an A is the k mac chem type an a cal esuant the 90th percentile of thi at the 90th percentile of this core normal distribution. t is known that the distribution of purchase 2 Suppose it is known customers entering a popular retail store amounts by custo is approximately normal with me deviation $20. What is the probability that a randomly selected customer spends less than $85 at this store? b. What is the probability that a randomly selected customer spends between $65 and $85 at this store?

Explanation / Answer

a.       Mean=75, SD=20

For X<85,

Find z-score ,

Z = 85-75/20 = 0.5

Now find p-value for z=0.5 Mean=75, SD=20

For X<85,

As per table, this comes to 0.6915

Subtract this from 1, which comes to 1-0.6915 = 0.3085 is the answer

b.      Mean=75, SD=20

For 65 =< X >= 85

z-score for 65,

z= 65-75/20 = -0.5

z-score for 85,

z=85-75/20= 0.5

p(z<0.5) = 0.6915

p(z< -0.5) = 0.3085

Subtract the above values, 0.6915-0.3085 = 0.3830 is the answer

c.       Mean=75, SD=20

For X>45,

Z = 45-75/20       = 1.5

P (z>1.5) = 0.9332

Subtract – 1-0.9332 = 0.0668 is the answer

d.      Mean=75, SD=20

For p > .75, find X

Subtract from 1, 1-.75 = .25

P(z<.25) = 0.5987

So, 0.5987 = X-75/20

i.e. 0.5987 * 20 = x-75

i.e. 11.974+75 = x

i.e. x= 75+12 (rounded off) = 87 dollars is the answer

e.      Mean =75, SD = 20

For p <0.8, find X

P (z<0.8) = 0.7881

So, 0.7881= x-75/20

Ie. 15.76 + 75 =x = 91 dollars is the answer

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