Q: In the same study, what is the probability that fewer than 5 out of 30 randomly surveyed individuals do NOT sanitize their hands after shopping?

$$P(X \u003c 5) = \text{binomcdf}(30, 0.32, 4) = 0.0171$$

Q: A recent poll found that 45% of adults believe the overall state of moral values is poor. For a random sample of 500 adults, what is the expected number (mean) and standard deviation of adults who believe this?

Mean: $$\mu = 500 \times 0.45 = 225$$, Standard deviation: $$\sigma = \sqrt{500 \times 0.45 \times 0.55} = 11.13$$

Q: In a city, 35% of murders are committed with a knife. In a random sample of 300 murders, what is the expected number of murders by knife and the standard deviation?

Expected: $$E(X) = 300 \times 0.35 = 105$$, Standard deviation: $$\sigma = \sqrt{300 \times 0.35 \times 0.65} = 8.36$$

Q: A clinical trial found that the drug Zylogan was 60% effective in reducing joint pain. In a sample of 250 adults, what is the probability that the drug was effective for exactly 155 of them?

$$P(X=155) = \text{binompdf}(250, 0.60, 155) = 0.041$$

Q: What is the probability that at most 130 out of 250 adults found relief from joint pain after taking Zylogan?

$$P(X \leq 130) = \text{binomcdf}(250, 0.60, 130) = 0.0002$$

Q: What is the probability that more than 200 out of 250 adults found relief from joint pain after taking Zylogan?

$$P(X \u003e 200) = 1 - \text{binomcdf}(250, 0.60, 200) = 0.0001$$

Question 1

A university study found that 32% of individuals do NOT sanitize their hands after shopping. If you survey 30 randomly selected individuals, what is the probability that exactly 7 of them do not sanitize their hands after shopping? Use the binomial probability formula.

Accepted Answer

$$P(X=7) = \binom{30}{7} (0.32)^7$ (0.68$$)^{23}$$ = 0.0938$$

Question 2

In the same study, what is the probability that fewer than 5 out of 30 randomly surveyed individuals do NOT sanitize their hands after shopping?

Accepted Answer

$$P(X \u003c 5) = 	ext{binomcdf}(30, 0.32, 4) = 0.0171$$

Question 3

Suppose you observe that fewer than 4 out of 30 individuals do NOT sanitize their hands after shopping. Would this be considered unusual? Why?

Accepted Answer

Yes, because $$P(X \u003c 4) = 0.0125 \u003c 0.05$$, which is considered unusual.

Question 4

A recent poll found that 45% of adults believe the overall state of moral values is poor. For a random sample of 500 adults, what is the expected number (mean) and standard deviation of adults who believe this?

Accepted Answer

Mean: $$\mu = 500 	imes 0.45 = 225$$, Standard deviation: $$\sigma = \sqrt{500 	imes 0.45 	imes 0.55} = 11.13$$

Question 5

Would it be unusual if 450 out of 500 adults surveyed believe that the overall state of moral values is poor? Use the mean and standard deviation from the previous question.

Accepted Answer

Yes, because $$450 is $$more than $$2$$ standard deviations above the mean ($$225 + 2 	imes 11.13 = 247.26$$), so it is unusual.

Question 6

In a city, 35% of murders are committed with a knife. In a random sample of 300 murders, what is the expected number of murders by knife and the standard deviation?

Accepted Answer

Expected: $$E(X) = 300 	imes 0.35 = 105$$, Standard deviation: $$\sigma = \sqrt{300 	imes 0.35 	imes 0.65} = 8.36$$

Question 7

Would it be unusual to observe 150 murders by knife in a random sample of 300 murders? Use the mean and standard deviation from the previous question.

Accepted Answer

Yes, because $$150 is $$much greater than $$E(X) + 2\sigma = 105 + 2 	imes 8.36 = 121.72$$, so it is unusual.

Question 8

A clinical trial found that the drug Zylogan was 60% effective in reducing joint pain. In a sample of 250 adults, what is the probability that the drug was effective for exactly 155 of them?

Accepted Answer

$$P(X=155) = 	ext{binompdf}(250, 0.60, 155) = 0.041$$

Question 9

What is the probability that at most 130 out of 250 adults found relief from joint pain after taking Zylogan?

Accepted Answer

$$P(X \leq 130) = 	ext{binomcdf}(250, 0.60, 130) = 0.0002$$

Question 10

What is the probability that more than 200 out of 250 adults found relief from joint pain after taking Zylogan?

Accepted Answer

$$P(X \u003e 200) = 1 - 	ext{binomcdf}(250, 0.60, 200) = 0.0001$$

Question 11

The lengths of human pregnancies are approximately normally distributed with mean $$\mu = 266$$ days and standard deviation $$\sigma = 16$$ days. What proportion of pregnancies last more than 270 days?

Accepted Answer

$$z = \frac{270 - 266}{16} = 0.25$$, $$P(Z \u003e 0.25) = 1 - 0.5987 = 0.4013$$

Question 12

What proportion of pregnancies last less than 250 days? (Mean $$= 266$$, SD $$= 16)$$

Accepted Answer

$$z = \frac{250 - 266}{16} = -1.0$$, $$P(Z \u003c -1.0) = 0.1587$$

Question 13

What proportion of pregnancies last between 240 and 280 days? (Mean $$= 266$$, SD $$= 16)$$

Accepted Answer

$$z_1$$ = \frac{240 - 266}{16} = -1.63$$, z_2$ = \frac{280 - 266}{16} = 0.88$$, $$P(-1.63 \u003c Z \u003c 0.88) = 0.759$$

Question 14

A 'very preterm' baby is one whose gestational period is less than 224 days. Are very preterm babies unusual? (Mean $$= 266$$, SD $$= 16)$$

Accepted Answer

Yes, because $$z = \frac{224 - 266}{16} = -2.63$$, $$P(Z \u003c -2.63) = 0.0043 \u003c 0.05$$, so it is unusual.

Question 15

If the mean time between eruptions of a geyser is 90 minutes with a standard deviation of 40 minutes, what is the probability that the time between eruptions is longer than 130 minutes?

Accepted Answer

$$z = \frac{130 - 90}{40} = 1.0$$, $$P(Z \u003e 1.0) = 1 - 0.8413 = 0.1587$$

Question 16

A random sample of 500 college guides found that 43% were accurate. What is the standard deviation of the sample proportion?

Accepted Answer

$$\sigma = \sqrt{\frac{0.43 	imes 0.57}{500}} = 0.022$$

Probability Distributions and Statistical Inference: Binomial and Normal Applications

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