According to Nielsen, adult Americans spend 2.35 hours per day watching television on a weekday, with a standard deviation of 1.93 hours. If a random sample of 40 adult Americans is obtained, determine the probability that a random sample of 40 adult Americans results in a mean time watching television on a weekday of between 2 and 3 hours.

0.8583

Step-by-step explanation:

Given the following :

Standard deviation (sd) = 1.93

Number of samples (s) = 40

Mean (m) = 2.35

x = 2 and 3

Firstly : find the z_score for both x values

Recall :

Z = (x - m) / (sd/√n)

For x = 2

z = (2 - 2.35) / (1.93/√40)

z = -0.35 / 0.3051597

z = 1.1469400

z = 1.15

For x =3

z = (3 - 2.35) / (1.93/√40)

z = 0.65 / 0.3051597

z = 2.1300322

z = 2.13

Therefore;

P(2<=m<=3) = P(-1.15<z<2.13)

P(z<2.13) - P(z<-1.15)

Using the z - table:

Z = - 1.1 on the vertical and 0.05 horizontal = 0.1251

Z = 2.1 vertically, 0.03 horizontal = 0.9834

0.9834 - 0.1251 = 0.8583

your answer would be 12