Question

According to a survey by Accountemps, 48% of executives believe that employees are most productive on Tuesdays. Suppose 230 executives are randomly surveyed.

Appendix A Statistical Tables

a. What is the probability that fewer than 103 of the executives believe employees are most productive on Tuesdays?
b. What is the probability that more than 115 of the executives believe employees are most productive on Tuesdays?
c. What is the probability that more than 94 of the executives believe employees are most productive on Tuesdays?

(Round the values of z to 2 decimal places. Round the intermediate values to 4 decimal places. Round your answers to 4 decimal places.)

Answer 1

Answer)

N = 230

P = 0.48

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 110.4

N*(1-p) = 119.6

Both the conditions are met so we can use standard normal z table to estimate the probability

Mean = n*p = 110.4

S.d = √{n*p*(1-p)} = 7.57680671523

Z = (x - mean)/s.d

A)

P(x<103)

By continuity correction

P(x<102.5)

Z = (102.5-110.4)/7.57680671523

Z = -1.04

From z table, P(z<-1.04) = 0.1492

B)

P(x>115)

By continuity correction

P(x>115.5)

Z = (115.5 - 110.4)/7.57680671523

Z = 0.67

From z table, P(z>0.67) = 0.2514

C)

P(x>94.5) by continuity correction

Z = -2.1

From z table, P(z>-2.1) = 0.9821