WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the normal distribution. The mean of the distribution is 15.0 minutes and the standard deviation is 3.5 minutes. What is the probability that a particular listener will tune in:

Complete question :

WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the normal distribution. The mean of the distribution is 15.0 minutes and the standard deviation is 3.5 minutes. What is the probability that a particular listener will tune in:

a. More than 20 minutes? b. For 20 minutes or less

0.0764 ; 0.9236

Step-by-step explanation:

Mean (m) = 15 minutes

Standard deviation (sd) = 3.5

Z - score :

Z = (x - m) / sd

A) more than 20 minutes

P(X > 20) ; x = 20

Z > (20 - 15) / 3.5 = 5/3.5 = 1.429

P(z > 1.429) = 0.5 - P(0<z<1.43)

Using the z-table: 1.43 = 0.4236

0.5 - 0.4236 = 0.0764

2) 20 minutes or less :

P(X <= 20) ; x = 20

Z < (20 - 15) / 3.5 = 5/3.5 = 1.429

P(z < 1.429) = 0.5 + P(0<z<1.43)

Using the z-table: 1.43 = 0.4236

0.5 + 0.4236 = 0.9236

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3