Question

According to the Centers for Disease Control 50% of practicing physicians are in the specialty area of primary care. Assuming that the same rate prevails, find the mean and standard deviation of the number of physicians specializing in primary care out of a current random selection of 545 medical graduates.
5.78 According to the Mendelian theory of inheritance of genes, offspring of a dihybrid cross of peas could be any of the four types: round-yellow (RY ), wrinkled-yellow (WY ), round-green (RG) and wrinkled-green (WG), and their probabilities are in the ratio 9:3:3:1.

(a) If X denotes the number of RY offspring from 130 such crosses, find the mean and standard deviation of X.
(b) If Y denotes the number of WG offspring from 85 such crosses, find the mean and standard deviation of Y.

Answer 1

Solution :

1) Let X represents the number of physicians specializing in primary care out of a current random selection of 545 medical graduates.

Given that, 50% of of practicing physicians are in the specialty area of primary care.

Hence, probability that a physician is in speciality are of primary care is 50/100 = 0.50

Let use consider "a physician who is specialist in primary care" as success. So, now we have only two mutually exclusive outcomes (success and failure).

Probability of success (p) = 0.50

Number of trials (n) = 545

Since, probability of success remains constant for each of the trials, number of trials are finite, we have only two mutually exclusive outcomes for each of the trials and outcomes are independent, therefore we can consider that X follows binomial distribution.

For a binomial distribution, the mean is given as follows :

Mean = np

Where, n is number of trials and p is probability of success.

We have, n = 545 and p = 0.50

Hence, mean = 545 × 0.50

mean = 272.5

The mean is 272.5.

For a binomial distribution, the standard deviation is given as follows:

The standard deviation is 11.6726.