Question

Almost three-fourths of US pecans are grown in the states of GA, NM, and TX. It is known that the average yield of pecans in Dona Ana County μ = 1625 lbs/acre with a standard deviation of σ = 200 lbs/acre. Are the following statements true or false.

If a RANDOM SAMPLE of 18 acres of land is taken, the probability of the average yield of these 18 acres being at most 1700 lbs/acre would be .9441.

If the average yield from a RANDOM SAMPLE of 18 acres of land is 1700 lbs/acre, this sample average would fall above the 95th percentile of all such sample means.

Answer #1

**ANSWER::**

Here average yield = = 1625 lbs.acre

Standard deviation = = 200 lbs.acre

**(a)**

ample size = n = 18

standard error = /sqrt(n) = 200/sqrt(18) = 47.14 lbs/acre

P( 1700) = NORMSDIST( 1700; 1625; 47.14)

z = (1700 - 1625)/47.14 = 1.591

so using NORMSDIST function or using z table

P( 1700) = P(Z < 1.591) = 0.9441

so the statement is true here.

**(b)**

The statement is false here. The sample average will fall below the 95th percentile of all such sample means.

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