Question

An adult female of species felis catus has given birth to five individuals. The birth weight of the first four young, in grams, are x = 99, 82, 92, 91.

(a) What is the mean weight of the first four young?

(b) What is the standard deviation of the weight of the first four young?

(c) Assuming the birth weights of felis catus are normally distributed, what is the probability that the fifth young has a birth weight of 90g or less? (This must be expressed as a number between 0 and 1)

Answer 1

(a) Mean of the first four could be obtained by adding the four ages and dividing by 4, as follows:

(b) Standard deviation (S.D) is found by taking the square root of the following fraction:

Numerator: Sum of squares of differences of each value from the mean

Denominator: Number of values

Thus we get:

(c) The probability that the fifth young has a weight less than 90g involves the following steps:

Step-1: Convert the value 90 into standard Z-value using the mean and standard deviation calculated in steps (a) and (b) using the below:

Thus we get,

Step-2: Assuming that the age of the 5th Young is X, now the probability of X<90 is equivant to probability of Z < -0.1655. By the virtue of Normal distribution is symmetric in nature, and that the total area under the normal curve is 1, probability of Z < -0.1655 is nothing but (1 - probability of Z < 0.1655). Probability of Z < 0.1655 can be easily obtained from standard normal distribution tables, which is 0.5675. hence the probability of Z<-0.1655 = probability of X < 90 = probability of the fifth young less than 90 = 1-0.5675 = 0.4325

(Note:- to obtain the probability from the standard normal distribution table, only the first two digits post the decimal point can be used. Look for the first digit in the column index in the left and the second digit in the row index on the top. hence in our case, 0.1655 was rounded off to 0.17, and the value intersecting 0.1 on the left and 0.07 on the top has been taken as the required probability. Refer the below image)