Question

1. a local radio station claims they have limited commercial breaks. let c = the number of commercials played during a randomly selected hour on this radio station. the probability distribution of c is shown in the table.

c	0	1	2	3	4	5

a. find ( p(c = 5) ).

b. find and interpret the mean of c.

c. find and interpret the standard deviation of c.

Explanation:

Step1: Sum known probabilities

$\sum P(C) = 0.05 + 0.15 + 0.20 + 0.25 + 0.25 = 0.90$

Step2: Calculate $P(C=5)$

$P(C=5) = 1 - 0.90 = 0.10$

Step3: Compute the mean $\mu$

$\mu = (0 \times 0.05) + (1 \times 0.15) + (2 \times 0.20) + (3 \times 0.25) + (4 \times 0.25) + (5 \times 0.10)$
$\mu = 0 + 0.15 + 0.40 + 0.75 + 1.00 + 0.50 = 2.8$
Interpretation: Over many randomly selected hours, the average number of commercials per hour is 2.8.

Step4: Compute variance $\sigma^2$

$\sigma^2 = (0-2.8)^2(0.05) + (1-2.8)^2(0.15) + (2-2.8)^2(0.20) + (3-2.8)^2(0.25) + (4-2.8)^2(0.25) + (5-2.8)^2(0.10)$
$\sigma^2 = (7.84 \times 0.05) + (3.24 \times 0.15) + (0.64 \times 0.20) + (0.04 \times 0.25) + (1.44 \times 0.25) + (4.84 \times 0.10)$
$\sigma^2 = 0.392 + 0.486 + 0.128 + 0.01 + 0.36 + 0.484 = 1.86$

Step5: Compute standard deviation $\sigma$

$\sigma = \sqrt{1.86} \approx 1.42$
Interpretation: The number of commercials in a randomly selected hour typically varies by about 1.42 from the mean of 2.8.

Answer:

a. $P(C=5)=0.10$
b. Mean of $C = 2.8$
c. Standard deviation of $C \approx 1.42$