Question

Question 12

Consider the following binomial experiment: A newspaper publisher claims that 52% of the people in a certain community read his newspaper. Doubting the assertion, a competitor randomly surveys 254 people in the community. Find the standard deviation of this experiment, if the newspaper publisher's claim is correct.

a) 8.2623

b) 7.9623

c) 7.7623

d) 7.6623

e) 7.5623

f) None of the above.

Question 13

Let Z be a standard normal variable. Find P(Z < 0.19).

a) 0.5398

b) 0.4602

c) 0.4247

d) 0.5753

e) 0.5793

f) None of the above.

Question 14

Let Z be a standard normal variable. Find P(0.50 < Z < 2.51).

a) 0.3072

b) 0.3023

c) 0.3112

d) 0.3025

e) 0.3176

f) None of the above.

Question 15

Let Z be a standard normal variable. Find the value of z if z satisfies P( -z < Z < z) = 0.3616.

a) -0.35

b) 0.40

c) 0.47

d) 0.30

e) 0.48

f) None of the above.

Answer 1

solution:-

Question 12 option b)7.9623
given p = 0.52 , n = 254
standard deviation = sqrt(n*p*(1-p))
= sqrt(254*0.52 * (1-0.52))
= 7.9623

question 13 option d)0.5753
p(z < 0.19) = 0.5753

question 14. option d)0.3025
p(0.50 < Z < 2.51) = P(Z < 2 .51) − P(Z < 0.50)
= 0.994 - 0.6915
= 0.3025

Question 15 option c) 0.47
p(-z < Z < z) = 0.3616
P(Z < 0.47) − P(Z < −0.47) = 0.6808 - 0.0.3192
= 0.3616