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Section 5.2 #11: The monthly utility bills in a city are normally distributed, with a mean of

$\$100$ and a standard deviation of

$\$12$ . Find the probability that a randomly selected utility bill is
a.) less than

$\$70$ ,
b.) between

$\$90$ and

$\$120$ , and
c.) more than

$\$140$ .
Solution: We are told

$\mu=100$ and

$\sigma=12$ . For part a.), we are asked to compute

$P(x \lt 70)$ . To do this, we need to find the

$z$ -score:

$z=\dfrac{x-\mu}{\sigma}=\frac{70-100}{12}=-2.5$ Now we see that

$P(x < 70)=P(z \lt -2.5) \stackrel{\mathrm{table}}{=}0.0062.$

For part b.), we are asked to compute

$P(90 < x < 120)$ . Find the relevant

$z$ -scores:

$\dfrac{90-100}{12} \lt z \lt \dfrac{120-100}{12},$ or equivalently

$-0.83 \lt z \lt 1.66.$ Now we see that

$\begin{array}{ll} P(90 \lt x \lt 120) &= P(-0.83 < z < 1.66) \\ &= P(z <1.66) - P(-0.83) \\ &\stackrel{\mathrm{table}}{=} 0.9515 - 0.2033 \\ &= 0.7482 \end{array}$ For part c.), we are asked to compute

$P(x > 140)$ . First convert this to a

$z$ -score:

$z = \dfrac{x-\mu}{\sigma}=\dfrac{140-100}{12} = 3.33.$ Now we see that

$P(x>140) = P(z>3.33) = 1-P(z<3.33) \stackrel{\mathrm{table}}{=} 1-0.9995=0.0005.$

Section 5.3 #14: Find the

$z$ -score that corresponds to the percentile

$P_{40}$ . (If the area is not in the table, use the nearest value or the average.)
Solution: This percentile is

$P_{40}=0.4$ . We must search through the table for a

$z$ -score that corresponds as close as possible to probability

$0.4$ . Notice the normal table says the probability for

$z=-0.25$ is

$0.4013$ and the probability for

$z=-0.26$ is

$0.3974$ . The

$z$ -score corresponding to

$P_{40}$ must lie between these values, so take their average:

$z=\dfrac{-0.25 + -0.26}{2}=-0.255.$

Section 5.3 #37: The annual per capita consumption of fresh apples (in pounds) in the United States can be approximated by a normal distribution, with a mean of 9.5 pounds and a standard deviation of 2.8 pounds.
a.) What is the smallest annual per capital consumption of apples that can be in the top 25% of consumptions?
b.) What is the largest annual per capita consumption of apples that can be in the bottom 15% of consumptions?
Solution: In this problem we have

$\mu=9.5$ and

$\sigma=2.8$ .

For part a.), we want to find the

$x$ -value corresponding to the "top 25% of consumptions". This means we should focus on

$P_{75}=0.75$ (not

$P_{25}$ -- that would be the lowest 25% of consumptions). The normal table suggests that the

$z$ -score should lie between

$0.67$ and

$0.68$ , so take their average to get

$z=\dfrac{0.67+0.68}{2}=0.675.$ Now we must find the

$x$ -value corresponding to this

$z$ -score. Recall that

$z=\dfrac{x-\mu}{\sigma}$ so algebra tells us that

$x=z \sigma+\mu$ . Calculate this value with the

$z$ -score we found:

$x=(0.675)(2.8)+9.5=11.39.$ This means that the smallest annual per capital consumption of apples that can be in the top 25% of consumptions is 11.39.

For part b.), we want to find the

$x$ -value corresponding to the "bottom 15% of consumptions". This means we should focus on

$P_{15}=0.15$ . The normal table suggests that the

$z$ -score should be between

$-1.04$ and

$-1.03$ , so we will take

$z$ to be their average:

$z=\dfrac{-1.04+(-1.03)}{2}=-1.035$ . To find

$x$ , calculate

$x=z\sigma+\mu=(-1.035)(2.8)+9.5=6.602.$

Section 5.4 #16: For a sample of

$n=100$ , find the probability of a sample mean being greater than

$24.3$ when

$\mu=24$ and

$\sigma=1.25$ .
Solution: Our mean of sample means is

$\mu_{\overline{x}}=\mu=24.3$ and our standard deviation of sample means is

$\sigma_{\overline{x}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.25}{\sqrt{100}}=0.125.$ We are asked to find the probability

$P(x > 24.3)$ . First find the relevant

$z$ -score:

$z=\dfrac{24.3-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{24.3-24}{0.125}=2.4$ So now we see

$P(x>24.3)=P(z>2.4)=1-P(z<2.4)\stackrel{\mathrm{table}}{=} 1-0.9918=0.0082.$

Section 5.4 #29: During a certain week, the mean price of gasoline in the New England region was

$\$3.796$ per gallon. A random sample of

$32$ gas stations is selected from this population. What is the probability that the mean price for the sample was between

$\$3.781$ and

$\$3.811$ that week? Assume

$\sigma=\$0.045$ .
Solution: We are told

$\mu=3.796$ ,

$n=32$ , and

$\sigma=0.045$ . The mean of sample means is

$\mu_{\overline{x}}=\mu=3.796$ and the standard deviation of sample means is

$\sigma_{\overline{x}} = \dfrac{\sigma}{\sqrt{n}}=\dfrac{0.045}{\sqrt{32}}=0.007954.$ We are asked to find the probability

$P(3.781 < x < 3.811)$ . First find the relevant

$z$ -scores:

$\dfrac{3.781-3.796}{0.007954} < z < \dfrac{3.811-3.796}{0.007954},$ or, simply,

$-1.88 < z < 1.88.$ Therefore we see

$\begin{array}{ll} P(3.781 < x < 3.811) &= P(-1.88 < z < 1.88) \\ &= P(z < 1.88) - P(z < -1.88) \\ &\stackrel{\mathrm{table}}{=} 0.9699 - 0.0301 \\ &=0.9398. \end{array}$