Question

The taste test for PTC (phenylthiourea) is a common class demonstration in the study of genetics. It is known that 70% of Canadians are “tasters” and 30% are “non-tasters.” Suppose a genetics class of size 20 does the test to see if they match the Canadian percentage of “tasters” and “non-tasters.” (Assume the assignment of students to classes constitutes a random process.)

a. What is the probability distribution of the random variable x, the number of “non-tasters” in the class?

b. Find P(3 < x < 9).

c. Find the mean of x.

d. Find the variance of

Answer 1

a.

x : Number of "non-tasters: in the class

Probability distribution of random variable x is binomial distribution.

Binomial Distribution

Probability distribution of X is Binomial If 'X' is the random variable representing the number of successes, the probability of getting ‘r’ successes and ‘n-r’ failures, in 'n' trails, ‘p’ probability of success ‘q’=(1-p) is given by the probability function

Mean of X = E(X) = np

Variance of X = npq

For the given case; Success is : Non-taster

Number of trails : Genetics class size:

Probability of success : p : Probability that a student is a 'non-taster' = 30/100 = 0.3

q = 1-p = 1-0.3 = 0.7

Probability that 'r' number of non-tasters in the class = P(x=r)

b.

P(3 < x< 9) = P(x=4) + P(x=5) + P(x=6) +P(x=7) + P(x=8)

x	P(x)	P(x)
4		0.1304
5		0.1789
6		0.1916
7		0.1643
8		0.1144

P(3 < x< 9) = P(x=4) + P(x=5) + P(x=6) +P(x=7) + P(x=8) = 0.1304+0.1789+0.1916+0.1643+0.1144=0.7796

P(3 < x< 9) = 0.7796

c. Mean of x = np = 20 x 0.3 = 6

Mean of x = 6

d. Variance of x = npq =20 x 0.3 x 0.7 = 4.2

Variance of x = 4.2