A farmer harvest hundreds of potatoes each day, where the weights are distribute

Question

A farmer harvest hundreds of potatoes each day, where the weights are distributed according to a Gaussian distribution. The morning he sampled 100 potatoes, obtained an average weight of 75 grams and discarded those which were at or below the 2.0 confidence interval,(2.3% of the crop) which he determined to be smaller than 65 grams.

(A) what was the standard deviation to which the average weight was measured?

(B) How many large potatoes did he harvest, where “large” are those weighing between 80 and 85 g?

(C) What fraction of his crop was “extra large” or weighing more than 85g?

(D) If the farmer samples another group of potatoes from the same batch, within what range of values should he expect the mean weight to be with a confidence level of 68.3%

Explanation / Answer

Answer to part a)

n = 100

Mean (M) = 75

P(x<65) = 0.0230

referring to the Z table we get :

Z = -2

.

The Z formula is :

Z = (x - M ) / (s)

.

On plugging the values we get

-2 = (65 - 75) / (s)

-2 * s = -10

s = 5

.

Answer to part b)

P(80 <x <85) = P(x < 85) - P(x < 80)

P(x < 85) = P(z < (85-75)/5) = P(z < 2) = 0.97725

P(x < 80) = P(z < (80-75)/5) = P(z < 1) = 0.84135

Thus P(80 < x < 85) = 0.97725 - 0.84135 = 0.1359

Thus number of large potatoes = 100 * 0.1359 = 13.59 or 14

Large potatoes = 14

.

Answer to part c)

P(x>85) = 1 - P(x < 85)

We got P(x < 85) = 0.97725

Thus , P (x > 85) = 1 - 0.97725 = 0.02275

Thus the fraction of his crop that was extra large is "0.02275"

.

Answer to part d)

The range of values with which we expect the mean wieght of potato to be , with confidence level 68.3% is :

M - s , M + s

75 -5 , 75 + 5

70 to 80

Thus we are 68.3% confident that the mean weight of potaoes for a new sample would lie between the values 70 gms and 80 gms

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