Question

1. a parents group was concerned that there were too many commercials shown during their childrens programs. they found that 20% of the times, there were 5 commercials, 25% of the time there were 6 commercials, 38% of the time there were 7 commercials, 10% of the time there were 8 commercials and 7% of the time, there were 9 commercials. find the mean, variance and standard deviation for the distribution.

interpret the mean:

the group though that the average number of commercials was 5. were they correct? why or why not?

Explanation:

Step1: Define variables and probabilities

Let $x$ = number of commercials, $P(x)$ = probability of $x$
$x_1=5, P(x_1)=0.20$; $x_2=6, P(x_2)=0.25$; $x_3=7, P(x_3)=0.38$; $x_4=8, P(x_4)=0.10$; $x_5=9, P(x_5)=0.07$

Step2: Calculate the mean

Multiply each $x$ by $P(x)$, sum results.

$$\begin{align*} \mu&=\sum xP(x)\\ &=(5\times0.20)+(6\times0.25)+(7\times0.38)+(8\times0.10)+(9\times0.07)\\ &=1 + 1.5 + 2.66 + 0.8 + 0.63 \end{align*}$$

Step3: Calculate variance $\sigma^2$

First find $\sum x^2P(x)$, subtract $\mu^2$.

$$\begin{align*} \sum x^2P(x)&=(5^2\times0.20)+(6^2\times0.25)+(7^2\times0.38)+(8^2\times0.10)+(9^2\times0.07)\\ &=(25\times0.20)+(36\times0.25)+(49\times0.38)+(64\times0.10)+(81\times0.07)\\ &=5 + 9 + 18.62 + 6.4 + 5.67\\ \sigma^2&=\sum x^2P(x)-\mu^2 \end{align*}$$

Step4: Calculate standard deviation $\sigma$

Take square root of variance.
$$\sigma=\sqrt{\sigma^2}$$

Answer:

Mean: $6.59$
$\sigma^2$: $1.5019$
$\sigma$: $\approx1.2255$

Interpret the mean:

On average, there are 6.59 commercials shown during children's programs.

Group correctness:

The group was not correct. The calculated mean number of commercials is 6.59, which is higher than their claimed average of 5.