# 4.11: Chapter 6 Solution (Practice + Homework)

- Page ID
- 51793

**1**.

ounces of water in a bottle

**3**.

2

**5**.

–4

**7**.

–2

**9**.

The mean becomes zero.

**11**.

\(z = 2\)

**13**.

\(z = 2.78\)

**15**.

\(x = 20\)

**17**.

\(x = 6.5\)

**19**.

\(x = 1\)

**21**.

\(x = 1.97\)

**23**.

\(z = –1.67\)

**25**.

\(z \approx –0.33\)

**27**.

0.67, right

**29**.

3.14, left

**31**.

about 68%

**33**.

about 4%

**35**.

between –5 and –1

**37**.

about 50%

**39**.

about 27%

**41**.

The lifetime of a Sunshine CD player measured in years.

**43**.

\(P(x < 1)\)

**45**.

Yes, because they are the same in a continuous distribution: \(P(x = 1) = 0\)

**47**.

\(1 – P(x < 3)\) or \(P(x > 3)\)

**49**.

\(1 – 0.543 = 0.457\)

**51**.

0.0013

**53**.

0.1186

**55**.

- Check student’s solution.
- 3, 0.1979

**57**.

0.154

**58**.

0.874

**59**.

0.693

**60**.

0.346

**61**.

0.110

**62**.

0.946

**63**.

0.071

**64**.

0.347

**66**.

c

**68**.

- Use the z-score formula. \(z = –0.5141\). The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.
- Use the z-score formula. \(z = 1.5424\). The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.
- \(\text{Height} = 79 + 3.5(3.89) = 92.615 \text{inches}\), which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely.

**70**.

- iv
- Kyle’s blood pressure is equal to \(125 + (1.75)(14) = 149.5\).

**72**.

Let \(X =\) an SAT math score and \(Y =\) an ACT math score.

- \(X = 720, \frac{720-520}{15}=1.74\) The exam score of 720 is 1.74 standard deviations above the mean of 520.
- \(z = 1.5\)

The math SAT score is \(520 + 1.5(115) \approx 692.5\). The exam score of 692.5 is 1.5 standard deviations above the mean of 520. - \(\frac{X-\mu}{\sigma}=\frac{700-514}{117} \approx 1.59\), the z-score for the SAT. \(\frac{Y-\mu}{\sigma}=\frac{30-21}{5.3} \approx 1.70\), the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).

**75**.

d

**77**.

- \(X \sim N(66, 2.5)\)
- \(0.5404\)
- No, the probability that an Asian male is over 72 inches tall is \(0.0082\)

**79**.

- \(X \sim N(36, 10)\)
- The probability that a person consumes more than 40% of their calories as fat is \(0.3446\).
- Approximately 25% of people consume less than 29.26% of their calories as fat.

**81**.

- \(X =\) number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
- \(X \sim N(3, 1.5)\)
- The probability that the child spends less than one hour a day unsupervised is \(0.0918\).
- The probability that a child spends over ten hours a day unsupervised is less than \(0.0001\).
- 2.21 hours

**83**.

- \(X =\) the distribution of the number of days a particular type of criminal trial will take
- \(X \sim N(21, 7)\)
- The probability that a randomly selected trial will last more than 24 days is \(0.3336\).
- 22.77

**85**.

- mean = 5.51, s = 2.15
- Check student's solution.
- Check student's solution.
- Check student's solution.
- \(X \sim N(5.51, 2.15)\)
- \(0.6029\)
- The cumulative frequency for less than 6.1 minutes is 0.64.
- The answers to part 6 and part 7 are not exactly the same, because the normal distribution is only an approximation to the real one.
- The answers to part 6 and part 7 are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
- The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.

**88**.

- \(n = 100; p = 0.1; q = 0.9\)
- \(\mu = np = (100)(0.10) = 10\)
- \(\sigma=\sqrt{n p q}=\sqrt{(100)(0.1)(0.9)}=3\)

- \(z=\pm 1 : x_{1}=\mu+\mathrm{z} \sigma=10+1(3)=13 \text { and } x_2=\mu-\mathrm{z} \sigma=10-1(3)=7.68 \%\) of the defective cars will fall between seven and 13.
- \(z=\pm 2 : x_{1}=\mu+\mathrm{z} \sigma=10+2(3)=16 \text { and } x_2=\mu-\mathrm{z} \sigma=10-2(3)=4.95 \%\) of the defective cars will fall between four and 16
- \(z=\pm 3 : x_{1}=\mu+\mathrm{z} \sigma=10+3(3)=19 \text { and } x_2=\mu-\mathrm{z} \sigma=10-3(3)=1.99 .7 \%\) of the defective cars will fall between one and 19.

**90.**

- \(n = 190; p = 1515 = 0.2; q = 0.8\)
- \(\mu = np = (190)(0.2) = 38\)
- \(\sigma=\sqrt{n p q}=\sqrt{(190)(0.2)(0.8)}=5.5136\)

- For this problem: \(P(34 < x < 54) = 0.7641\)
- For this problem: \(P(54 < x < 64) = 0.0018\)
- For this problem: \(P(x > 64) = 0.0000012\) (approximately 0)

**92**.

- 24.5
- 3.5
- Yes
- 0.67

**93**.

- 63
- 2.5
- Yes
- 0.88

**94**.

0.02

**95**.

0.37

**96**.

0.50