Question

the body temperatures of adults are normally distributed with a mean of 98.6 degrees Fahrenheit and a standard deviation of 0.60 degrees Fahrenheit if 36 adults are randomly selected find the probability that their mean body temperature is greater than 98.4 degrees Fahrenheit

Answer 1

Let X be the random variable denoting the body

temperature of adults.

X ~ N(98.6, 0.6) i.e. (X - 98.6)/0.6 ~ N(0,1)

Let M be the mean body weight of 36 adults chosen at

random.

Thus, E(M) = 98.6, s.d.(M) = 0.6.

The probability that mean body temperature is greater than

98.4 degrees = P(M > 98.4)

= P[(M - 98.6)/0.6 > (98.4 - 98.6)/0.6]

= P[(M - 98.6)/0.6 > - 0.3333]

= 1 - P[(M - 98.6)/0.6 <= -0.3333] = 1 - $\Phi$ (-0.3333)

= 1 - 0.3695 = 0.6305. (Ans).