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