Biology and Environmental Science Many backyard gardeners prefer Silver Queen Hy

Question

Biology and Environmental Science Many backyard gardeners prefer Silver Queen Hybrid corn. This late-season variety is very sweet and has tender, white kernels. In some .locations in the Northeast, gardeners have trouble harvesting this variety because of its longer growing time. The temperature of the soil should be at least 65degreeF before planting, and the growing time is approximately normal with mean 92 days and standard deviation 5 days. What is the probability that a randomly selected seed will mature in less than 90 days? What is the probability that a randomly selected seed will mature in between 95 and 100 days? Suppose a row in a backyard garden contains 12 plants, by the 95th day? Find a value h such that 99% of all plants are ready to be harvested within h days.

Explanation / Answer

WE HAVE BEEN GIVEN THE MEAN = 92

STANDARD DEVIATION = 5

THE DISTRIBUTION IS NORMAL AND THE Z VALUE WILL BE CALCULATED

Z SCORE = (X-MEAN)/STANDARD DEVIATION

A) P(X<90) =

For x = 90, the z-value z = (90 - 92) / 5 = -0.4

Hence P(x < 90) = P(z < -0.4), now from the z table we will take the value of z score = -0.4

And that value will be the probability required.

= [area to the left of -0.4] = 0.3446

B)

) For x = 95 , z = (95 - 92) / 5 = 0.6 and for x = 100, z = (100 - 92) / 5 = 1.6

Hence P(95 < x < 100) = P(0.6 < z < 1.6) = [area to the left of z = 1.6] - [area to the left of 0.6]

= 0.9452 - 0.7257 = 0.2195

C) FROM PART B THE PROBABILITY THAT PLANT WILL BE READY BY 95TH DAY = 0.7257

THEREFORE WE HAVE N = 12

P(X=4) = 12C4*(0.7257)^4*(0.2743)^8 = 0.0043

D) THE FORMULA TO BE USED =

X = MEAN+ Z*STANDARD DEVIATION

Z SCORE FOR 99% = 2.57

THEREFORE X = 92+2.57*5 = 104.85 = 105

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