Question

Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 grams. Determine the following. You must show your work or explain the reasoning or process you used to arrive at your answers. Please circle or highlight your final answers.
a. If a kitten weighs 99 grams at birth, what percentile of the weight distribution is it in?
b. How heavy must a kitten be, at a minimum, to be in the 90 th percentile (i.e. 90% of kittens weigh less than this amount)?

Answer 1

(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.

Weight distribution of the kitten

In a normal distribution curve;

• 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%

• 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
• 1 standard deviation (d) above the mean (M), (M + d) is at 84%
• 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

M - 2d = 125 g - 2(15g) = 95 g

M - d = 125 g - 15 g = 110 g

95 g is at 2% and 110 g is at 16%

(16% - 2%) = 14%

(110 - 95) = 15 g

14% / 15g = 0.93%/g

From 95 g to 99 g:

99 g - 95 g  = 4 g

4g x 0.93%/g = 3.72%

99 g will be at:

(2% + 3.72%) = 5.72%

Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

Weight of the kitten in the 90th percentile

M + d = 125 + 15 = 140 g      (at 84%)

M + 2d = 125 + 2(15) = 155 g   ( at 98%)

155 g - 140 g = 15 g

14% / 15g = 0.93%/g

84% + x(0.93%/g) = 90%

84 + 0.93x = 90

0.93x = 6

x = 6.45 g

weight of a kitten in 90th percentile = 140 g + 6.45 g  = 146.45 g

Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

Learn more about standard deviation here: brainly.com/question/475676

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