Question

Consider a symmetric distribution which has two modes, a mean of 68 and standard deviation of 6. Approximately what percentage of data lie between 56 and 80?

A)95%

B)75%

C)5%

D)13.5%

E)68%

Answer 1

Let X be the random variable denoting the distribution. We can assume Normal distribution without loss of generality.

Thus, X ~ N(68, 6) i.e. Z = (X - 68)/6 ~ N(0,1)

P(56 < X < 80) = P[(56 - 68)/6 < (X - 68)/6 < (80 - 68)/6] = P[ -2 < Z < 2] = $\Phi$(2) - $\Phi$(-2) = 0.9772 - 0.0228 = 0.9544

[$\Phi$(.) is the cdf of N(0,1)].

Hence, 95.44% of observations will lie between 56 and 80. Thus, Option (A) is the correct choice. (Ans).