Weights of female cats of a certain breed are normally distributed with mean 4.1

Question

Weights of female cats of a certain breed are normally distributed with mean 4.1 kg and standard deviation 0.6 kg. a. What proportion of female cats have weights between 3.7 and 4.4 kg? b. A certain female cat has a weight that is 0.5 standard deviations above the mean. What proportion of female cats are heavier than this one? c. How heavy is a female cat whose weight is on the 80th percentile? d. A female cat is chosen at random. What is the probability that she weighs more than 4.5 kg? e. Six female cats are chosen at random. What is the probability that exactly one of them weighs more than 4.5 kg?

Explanation / Answer

Given:

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Answer to part a)

P(3.7 < x < 4.4) = P(x < 4.4) - P(x < 3.7)

P(x < 4.4)

z = (4.4 - 4.1) / 0.6 = 0.5

P(x < 4.4) = P(z < 0.5) = 0.6915

.

P(x < 37)

z = (37. -4.1) / 0.6 = -0.67

P(x<3.7) = P(z < -0.67) = 0.2514

.

P(3.7 < x < 4.4) = 0.6915 - 0.2514 = 0.4401

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Answer to part b)

x = 4.1 + 0.5 = 4.6

P(x > 4.6) = 1 - P(x < 4.6)

P(x < 4.6)

Z = (4.6 - 4.1)/0.6 = 0.833

P(x < 4.6) = P(z < 0.833) = 0.7977

P(x > 4.6) = 1 - 0.7977 = 0.2023

Thus proportion of female cats who are heavier than 4.6 are 0.2023

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Answer to part c)

We need to find x for p = 0.80 ( or 80th percentile)

P(X < x) = 0.80

we need to find the Z vlaue for this area value from the Z table

we get Z = 0.8416

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Now we make use of the Z formula to get the value of weight (x) corresponding to 80th percentile

Z = (x - M ) / SD

0.8416 = (x - 4.1) / 0.6

x = 0.8416 *0.6 + 4.1 = 4.60496

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Answer to part d)

P(x > 4.5) = 1 - P(x < 4.5)

For P(X < 4.5)

z = (4.5 -4.1)/0.6 = 0.6667

P(x < 4.5) = P(Z < 0.6667) = 0.7475

Thus P(x > 4.5) = 1 - 0.7475 = 0.2525

.

Answer to part e)

For this part we make use of binomial probability

The formula of binomial probability is as follows:

x = number of success for which probability value needs to be calculated

n = total number of trials

p = probability of success, that is getting a female cat with weight above 4.5kg

P(X=x) = nCx * (p)^x*(1-p)^(n-x)

P(X=1) = 6C1 * (0.2525)^1 * (0.7475)^5

P(X=1) = 0.3536

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