Question

52. According to a government study among adults in the 25- to 34-year age group, the
mean amount spent per year on reading and entertainment is $1,994. Assume that the
distribution of the amounts spent follows the normal distribution with a standard deviation
of $450.
a. What percent of the adults spend more than $2,500 per year on reading and
entertainment?
b. What percent spend between $2,500 and $3,000 per year on reading and
entertainment?
c. What percent spend less than $1,000 per year on reading and entertainment?

Answer 1

Answer on Question #59183 – Math – Statistics and Probability

According to a government study among adults in the 25- to 34-year age group, the mean amount spent per year on reading and entertainment is $1,994. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $450.

Question

a. What percent of the adults spend more than $2,500 per year on reading and entertainment?

Solution

$$\begin{array}{l} \mu = 1994 \\ \sigma = 450 \\ x = 2500 \\ z = \frac{x - \mu}{\sigma} = \frac{2500 - 1994}{450} = 1.12(4) \end{array}$$

We will use the z-table below to calculate the percent.

So, from the z-table, 0.8888 of people spend less or equal 2500$ per year on reading.

More than 2500$ spend 1 – 0.8888 = 0.1112, that is, about 11% of adults.

Tables of the Normal Distribution

Probability Content from -oo to Z

Answer: $11\%$ .

Question

b. What percent spend between $2,500 and $3,000 per year on reading and entertainment?

Solution

We shall use the z-table above to calculate the percent.

$$p = p \left(z _ {3 0 0 0}\right) - p \left(z _ {2 5 0 0}\right)$$ $$p \left(z _ {2 5 0 0}\right) = 0. 8 8 8 8$$ $$z _ {3 0 0 0} = \frac {x - \mu}{\sigma} = \frac {3 0 0 0 - 1 9 9 4}{4 5 0} = 2. 2 4$$ $$p \left(z _ {3 0 0 0}\right) = p (2. 2 4) = 0. 9 8 7 5$$

$p = 0.9875 - 0.8888 = 0.0987 \approx 0.1$, that is, about $10\%$ of adults.

Answer: $10\%$

Question

c. What percent spend less than $1,000 per year on reading and entertainment?

Solution

This percent is equal to percent of people that spend more than $\mu + (\mu - 1000) = 2988$ .

$$p = 1 - p \left(z _ {2 9 8 8}\right)$$ $$z _ {2 9 8 8} = \frac {x - \mu}{\sigma} = \frac {2 9 8 8 - 1 9 9 4}{4 5 0} = 2. 2 1$$ $$p \left(z _ {2 9 8 8}\right) = p (2. 2 1) = 0. 9 8 6 4$$

$p = 1 - 0.9864 = 0.0136$, that is, about $1\%$ .

Answer: $1\%$ .

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