Question

The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.15 (a) If a man is chosen at random from the population, find the probability that he will be more than 76 inches tall. (b) If two men are chosen at random from the population, find the probability that (i) both of them will be more than 76 inches tall; (ii) their mean height will be more than 76 inches.

Answer 1

Answer:

a) P(Y > 76) = 0.0122

b) i) P(both of them will be more than 76 inches tall) = 0.00015

   ii) P(Y > 76) = 0.0007

Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

                    the probability that he will be more than 76 inches tall.

              (b) If two men are chosen at random from the population, find

                    the probability that

                    (i) both of them will be more than 76 inches tall;

                    (ii) their mean height will be more than 76 inches.

Proof -

a)

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( $\frac{( Y- mean)}{S.D}$) > $\frac{( 76- mean)}{S.D}$)

                 = P(Z >  $\frac{( 76- mean)}{S.D}$)

                 = P(Z > $\frac{76 - 69.7}{2.8}$)

                 = P(Z > 2.25)

                 = 1 - P(Z  ≤ 2.25)

                 = 0.0122

⇒P(Y > 76) = 0.0122

b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

$\frac{S.D}{\sqrt{N} }$ = 1.979899,

Now,

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( $\frac{( Y- mean)}{\frac{S.D}{\sqrt{N} } }$)) > $\frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } }$)

                 = P(Z > $\frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } }$)

                 = P(Z > $\frac{( 76- 69.7)}{1.979899 }$))

                 = P(Z > 3.182)

                 = 1 - P(Z ≤ 3.182)

                 = 0.0007

⇒P(Y > 76) = 0.0007