Question

1. Suppose that in a population of American adult coffee drinkers the change in heart rate ten minutes after drinking 6 ounces of coffee is normally distributed with a mean increase of 3.6 beats per minute and a standard deviation of 3.2 beats per minute. A negative change indicates a decrease in heart rate.
   1. What proportion of individuals in the population have a decrease in heart rate after drinking 6 ounces of coffee?
2. 2. The middle 80 percent of individuals have changes in heart rate between what two values?
3. 3. Find the probability that mean increase in heart rate with a sample of 20 subjects will exceed 6.8.

Answer 1

Mean, = 3.6

Standard deviation, = 3.2

(a) The proportion of individuals in the population have a decrease in heart rate = P(X < 0)

= P{Z < (0 - 3.6)/3.2}

= P(Z < -1.125)

= 0.1303

(b) Corresponding to middle 80 percent heart rate changes, the z value range is -1.2845 to 1.2845

Thus, the lower limit = 3.6 - 1.2845*3.2 = -0.51

Upper limit = 3.6 + 1.2845*3.2 = 7.71

(c) For sample of 20 subjects, the standard error = 3.2/√20 = 0.7155

The required probability = P{Z > (6.8 - 3.6)/0.7155}

= P(Z > 4.472)

= 0.00001