Weights of female cats of a certain breed are normally distributed with mean 4.1 kg and standard deviation 0.6 kg.
. 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?
(a) For 3.7, z = (3.7-4.1)/0.6=-0.67 . For 4.4, z = (4.4-4.1)/-0.6=0.50.
The area between z= ?0.67 and z= 0.50 is 0.6915 ? 0.2514 = 0.4401.
(b) The area to the right of z= 0.50 is 0.3085. Therefore approximately 30.85% of cats are heavier than this one.
(c) The z-score of the 80th percentile is ? 0.84.
The 80th percentile is therefore ? 4.1 + 0.84(0.6) = 4.604.
(d) z = (4.5-4.1)/-0.6=0.67 . The area to the right of z= 0.67 is 0.2514.
(e) Let X be the number of cats that weigh more than 4.5 kg. Using part (d), the probability that a cat weighs more than 4.5 kg is 0.2514. Therefore X~Bin(6, 0.2514).
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