Question

In: Statistics and Probability

Phenylthiocarbamide, or PTC, tastes very bitter to those who can taste it. However, about 30% of...

  1. Phenylthiocarbamide, or PTC, tastes very bitter to those who can taste it. However, about 30% of people cannot taste it, due to a recessive allele in their DNA. Assume that the proportion of people with this genetic trait is exactly 30%. A researcher administers PTC to 25 randomly- selected subjects. Let X be the number of these subjects who cannot taste PTC.

    1. What is the probability distribution of X?

    2. Calculate:

      (i) P(5≤X≤11)     (ii) the mean and standard deviation of X.   (iii) P( X > 9)

3. If only 2 of the subjects could not PTC, would this be an unusual sample? Justify your answer.

Solutions

Expert Solution

It is a binomial distribution.

i) P(5 < X < 11) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)

                        = 25C5 * (0.3)^5 * (0.7)^20 + 25C6 * (0.3)^6 * (0.7)^19 + 25C7 * (0.3)^7 * (0.7)^18 + 25C8 * (0.3)^8 * (0.7)^17 + 25C9 * (0.3)^9 * (0.7)^16 + 25C10 * (0.3)^10 * (0.7)^15 + 25C11 * (0.3)^11 * (0.7)^14 = 0.8653

ii) mean = n * p = 25 * 0.3 = 7.5

standard deviation = sqrt(np(1 - p))

                             = sqrt(25 * 0.3 * 0.7) = 2.2913

iii) P(X > 9) = 1 - P(X < 9)

                  = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + p(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9))

                 = 1 - (25C0 * (0.3)^0 * (0.7)^25 + 25C1 * (0.3)^1 * (0.7)^24 + 25C2 * (0.3)^2 * (0.7)^23 + 25C3 * (0.3)^3 * (0.7)^22 + 25C4 * (0.3)^4 * (0.7)^21 + 25C5 * (0.3)^5 * (0.7)^20 + 25C6 * (0.3)^6 * (0.7)^19 + 25C7 * (0.3)^7 * (0.7)^18 + 25C8 * (0.3)^8 * (0.7)^17 + 25C9 * (0.3)^9 * (0.7)^16)

               = 1 - 0.8106 = 0.1894

3) P(X = 2) = 25C2 * (0.3)^2 * (0.7)^23 = 0.0074

Since the probability is less than 0.05, so it is unusual.


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