Kindly show how to do in Excel: Suppose that the weight of a typical American fe

Question

Kindly show how to do in Excel: Suppose that the weight of a typical American female follows a normal distribution with = 140 lb. and = 20 lb. Also, suppose 80.15% of all American females weigh more than I weigh. What fractions of American females weigh more than 165 pounds? How much do I weigh? If I weighed 20 pounds more than I do, what percentile would I be in? A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation 0.1. An item is considered to be defective if its weight is less than 9.8 pounds or greater than 10.2 pounds. Suppose that these items are currently produced in batches of 1000 units. Find the probability that at most 5% of the items in a given batch will be defective Find the probability that at least 85% of these items in a given batch will be acceptable.

Explanation / Answer

here as it is given that for aa typical American female

= 140 lb. and = 20 lb

Here in excel we use direct formulas

(a) Here 80.15% of all American females weigh more than I weigh. So, percentile of I weigh is 1 - 0.8015 = 0.1985

Here my weight is = NORMDIST(0.1985; 140 ; 20) = 123.06 lb

(b) Here fraction of American females weigh more than 165 pounds

Z = (165 - 140)/20 = 1.25

so, the probability of a American female weigh more than 165 pounds = 1 - NORMDIST(165,140,20,TRUE) = 1 - 0.89435 = 0.10565

(c) If my wight is 20 pounds more than it means my weight is 123.06 + 20 = 143.06 pounds

so the percentile is = NORMDIST(165,140,20,TRUE) = 0.5608

Question 2

Here mean weight = 10 pounds

standard deviation = 0.1 pounds

Defective item would be when weight is less than 9.8 pounds or greater than 10.2 pounds.

Pr(A batch is defective) = NORMDIST(10.2 pounds ; 10 pounds ; 0.1 pounds, true) - NORMDIST(9.8 pounds ; 10 pounds ; 0.1 pounds, true)

= 0.97725 - 0.02275 = 0.9545

Pr(Defective) = 1 - 0.9545 = 0.0455

Batch is in 1000 units. Here

Expected number of defective items in batch of 1000 = 1000 * 0.0455 = 45.5 items

Standard deviation = sqrt (1000 * 0.0455 * 0.9545) = 6.59 items

so here the proportion of defective items would have normal distribution

with proportion p = 0.0455

standard error of proportion = 0.00659

(a) Pr(at most 5% of the items in a given batch is defective) = 1 - NORMDIST (0.05 ; 0.0455 ; 0.00659, true) = 1 - 0.7526 = 0.2474

(b) Pr(at least 85% items are acceptable) = Pr(less than 15% items are acceptable)

= NORMDIST(0.15; 0.0455; 0.00659 ; true) = 1

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