Question

Thirty eight percent of US adults will be the victim of a crime in their lifetime based on n = 1014 US adults.

What is the % chance that another sample of US adults who have been the victim of a crime in their lifetime will be P ( p > 41%)?

What is the % chance that another sample of US adults who have been the victim of a crime in their lifetime will be P ( p < 38%)?

What is the % chance that another sample of US adults who have been the victim of a crime in their lifetime will be P ( 34% < p < 37%)?

Answer 1

Answer)

N = 1014

P = 0.38 (38%)

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 = 385.32

N*(1-p) = 628.68

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

z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

A)

P(x>0.41)

Observed P = 0.41

Claimed P = 0.38

N = 1014

After substitution

Z = 1.97

From z table, P(z>1.97) = 0.0244

B)

P(x<0.38)

Observed P = 0.38

Rest is same

Z = 0

From z table, P(z<0) = 0.5

C)

P(0.34<x<0.37) = P(x<0.37) - P(x<0.34)

P(x<0.37)

Z = -0.66

From z table, P(z<-0.66) = 0.2546

P(x<0.34)

Z = -2.62

From z table, p(z<-2.62) = 0.0044

Required answer is 0.2546 - 0.0044 = 0.2502