A grocery store manager is told by a wholesaler that the apples in a large shipm

Question

A grocery store manager is told by a wholesaler that the apples in a large shipment have a mean weight of 5 ounces and a standard deviation of 1 ounce. a) If the distribution of the weight is normal and a random sample of 10 b) If the distribution of the weight is normal, and the probability that the c) The manager is going to randomly select 100 apples and will decide to apples is selected, what is the probability that the sample total weight is between 45 and 55? average weight for n randomly selected apples is greater than 5.6 is 0.0505, find the sample size of the random sample. return the shipment if the mean weight of the sample is less than 4.8 ounces. What is the probability that the shipment will be returned? (Be careful, in this part, and the following one, I am not stating that the distribution is normal.) d) The probability that the total weight for 40 randomly selected apples is greater than a certain amount is 0.4483. Find this total weight.

Explanation / Answer

a) P(45/10 < X < 55/10)

= P(4.5 < X < 5.5)

= P((4.5 - 5)/1 < (X - mean)/sd < (5.5 - 5)/1)

= P(-0.5 < Z < 0.5)

= P(Z < 0.5) - P(Z < -0.5)

= 0.6915 - 0.3085

= 0.383

b) P(X > 5.6) = 0.0505

or, P(Z > (5.6 - 5)/(1/sqrt(n))) = 0.0505

or, P(Z < 0.6/(1/sqrt(n))) = 0.9495

or, 0.6/(1/sqrt(n)) = 1.64

or, sqrt(n) = 1.64/0.6

or, sqrt(n) = 2.73

or, n = 7

c) P(X < 4.8)

= P(Z < (4.8 - 5)/1)

= P(Z < -0.2)

= 0.4207

d) P(X > x) = 0.4483

P(Z > (x - 5)/1) = 0.4483

or, P(Z < (x - 5)) = 0.5517

or, x - 5 = 0.13

or, x = 5.13

So the total weight = 5.13 * 40 = 205.2

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