In a plantation of sunflowers, the height of plants tends to follow a normal dis

Question

In a plantation of sunflowers, the height of plants tends to follow a normal distribution, with a mean of 2.47 m and a standard deviation of 0.28 m.

e) A certain species of crawling insect is known to favour the roots of the variety of sunflower grown in this plantation. If 10 of these insects arrive in the plantation and each one selects a different sunflower, and it is assumed that they are unable to discern the heights of the plants from their position on the ground, then what is the probability, to the nearest %, that fewer than 3/4 of these insects will choose a sunflower with a height between 200 and 300 cm?

Explanation / Answer

First we find out the probability of choosing a plant whose height is between 200 and 300 cms.

Data given is:

Mean m = 2.47 m = 200 cm

Standard deviation S = 0.28 = 28 cm

At X = 200 cm, we have:

z = (X-m)/S = (200-247)/28 = -1.678

So,

p1 = P(X < 200) = P(z < -1.678) = 0.0467

At X = 300 cm, we have:

z = (X-m)/S = (300-247)/28 = 1.893

So,

p2 = P(X < 300) = P(z < 1.893) = 0.9708

So the required probability is:

p = (p2-p1) = 0.9708-0.0467 = 0.9241

Now we have calculated the probability of randomyl choosing a plant whose height is within the interval given.

Now this is a case of Binomial distribution with the following parameters:

n = 10, p = 0.9241

X = number of insects who select a plant within the given height

Fewer than 3/4 of 10, means fewer than 7.5, which means that 7 or fewer.

So we need to find the probability P( X <= 7)

Now,

P(X <= 7) = 1 - P(X>7)

P(X>7) = P(X=8) + P(X=9) + P(X=10)

P(X=8) = 10C8*(p^8)*((1-p)^(10-8)) = 10C8*(0.9241^8)*((1-0.9241)^(10-8)) = 0.138

P(X=9) = 10C9*(p^9)*((1-p)^(10-9)) = 10C9*(0.9241^9)*((1-0.9241)^(10-9)) = 0.373

P(X=10) = 10C10*(p^10)*((1-p)^(10-10)) = 10C10*(0.9241^10)*((1-0.9241)^(10-10)) = 0.454

So,

P(X>7) = 0.138 + 0.373 + 0.454 = 0.965

So,

P(X <= 7) = 1 - 0.965 = 0.035

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