Question

Playing video and computer games is very popular. Over 70% of households play such games. Of those individuals who play video and computer games, 18% are under 18 years old, 53% are 18-59 years old, and 29% are over 59 years old. For a sample of 600 people who play these games, what is the probability that fewer than 100 will be under 18 years of age? For a sample of 800 people who play these games, what is the probability that 200 or more will be over 59 years of age?

Answer 1

Answer:

(a) Probability that fewer than 100 will be under 18 years of age is 0.19655.

(b) Probability that 200 or more will be over 59 years of age is 0.00449.

Step-by-step explanation:

We are given that over 70% of households play such games. Of those individuals who play video and computer games, 18% are under 18 years old, 53% are 18-59 years old, and 29% are over 59 years old.

(a) A sample of 600 people is selected and we have to find the probability that fewer than 100 will be under 18 years of age.

The z score probability distribution for sample proportion is given by;

                          Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of people =  $\frac{100}{600}$  = 16.67%

            p = population proportion of people under 18 years old = 18%

            n = sample of people = 600

Now, probability that fewer than 100 will be under 18 years of age is given by = P( $\hat p$ < 0.167)

P( $\hat p$ < 0.167) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\frac{0.167-0.18}{\sqrt{\frac{0.167(1-0.167)}{600} } }$ ) = P(Z < -0.854) = 1 - P(Z $\leq$ 0.854)

                                                                   = 1 - 0.80345 = 0.19655

The above probability is calculated by looking at the value of x = 0.854 in the z table which will lie between x = 0.85 and x = 0.86.

(b) A sample of 800 people is selected and we have to find the probability that 200 or more will be over 59 years of age.

The z score probability distribution for sample proportion is given by;

                          Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of people =  $\frac{200}{800}$  = 25%

            p = population proportion of people over 59 years old = 29%

            n = sample of people = 800

Now, probability that 200 or more will be over 59 years of age is given by = P( $\hat p$ $\geq$ 0.25)

P( $\hat p$ $\geq$ 0.25) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\geq$ $\frac{0.25-0.29}{\sqrt{\frac{0.25(1-0.25)}{800} } }$ ) = P(Z $\geq$ -2.613) = P(Z $\leq$ 2.613)

                                                                = 1 - 0.99551 = 0.00449

The above probability is calculated by looking at the value of x = 2.613 in the z table which will lie between x = 2.61 and x = 2.62.