Question 6: In a plantation of sunflowers, the height of plants tends to follow

Question

Question 6: 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. a) Calculate, to the nearest%, the probability that a randomly-selected sunflower from this plantation b) Calculate, to the nearest %, the proportion of sunflowers from this plantation that are taller than 300 c) Calculate,to the nearest %, the probability that a randomly-selected sunflower from this plantation has d) It has been decided that the shortest 10% of sunflowers in the plantation are to be harvested for floral measures 2.47 m in height, using a measuring stick that is precise to the nearest cm? cm in height. a height between 200 and 300 cm. arrangements. Calculate, to the nearest cm, the minimum height for a sunflower to be harvested for this purpose. A certain species of crawling insect is known to favour the roots of the variety of sunflower grown in this plantation 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 00, that fewer than 3,4 of these insects will choose a sunflower with a height between 200 and 300 cm? e) . If 1 0 of these insects arrive in the plantation and each one selects a different sunflower ,

Explanation / Answer

(a)

Data given is as follows:

mean m = 2.47

Standard deviation S = 0.28

Here since it is given that the precision of the stick is upto the nearest cm, so we need to find out the probability that a plant has height between 2.465 m and 2.474 m

At X = 2.465, we have:

z = (X-m)/S = (2.465-2.47)/0.28 = -0.0178

The corresponding p-value is:

p1 = 0.4929

At X = 2.474, we have:

z = (X-m)/S = (2.474-2.47)/0.28 = 0.0142

The corresponding p-value is:

p2 = 0.5057

The required probability = p2-p1 = 0.5057-0.4929 = 0.0128 = 1.28%

(b)

At X = 300 cm = 3 m, we have:

z = (X-m)/S = (3-2.47)/0.28 = 1.893

The corresponding p-value is:

P(X > 300 cm = P(z > 1.893) = 0.0292 = 2.92% = 3%

(c)

At X = 200 cm = 2 m, we have:

z = (X-m)/S = (2-2.47)/0.28 = -1.678

The corresponding p-value is:

p1 = 0.0467

At X = 300 cm = 3 m, we have:

z = (X-m)/S = (3-2.47)/0.28 = 1.893

The corresponding p-value is:

p2 = 0.9708

The required probability = p2-p1 = 0.9708-0.0467 = 0.9241 = 92.41% = 92%

(d)

In this case you have to calculate maximum height, because minium height has no meaning here.

At p = 0.10, we have:

z = -1.28

So,

X = S*z + m = 0.28*(-1.280) + 2.47 = 2.116 m = 211.6 cm = 212 cm

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