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Question
5.1d.a.4: normal distributions 1
1 suppose two sets of test scores have the same mean, but different standard deviations, \\(\sigma_1\\) and \\(\sigma_2\\), with \\(\sigma_2 > \sigma_1\\). which statement best describes the variability of these data sets?
- data set one has the greater variability.
- the variability will be the same for each data set.
- data set two has the greater variability.
- no conclusion can be made regarding the variability of either set.
2 the job satisfaction rating at a company is approximately normally distributed with a mean of 12. about 95% of the scores are between 8 and 16. what is the standard deviation of this distribution? justify your answer.
3 the heights of women in the united states are normally distributed with a mean of 64 inches and a standard deviation of 2.75 inches. the percent of women whose heights are between 64 and 69.5 inches, to the nearest whole percent, is
- 6
- 68
- 48
- 95
4 a manufacturer claims that the number of ounces of a beverage dispensed by one of its automatic dispensers is normally distributed with a mean of 8.0 ounces and a standard deviation of 0.04 ounces. to the nearest tenth of a percent, what percent of the cups filled by this company’s dispenser will contain between 7.9 and 8.11 ounces?
- 99.5
- 99.1
- 99.4
- 97.6
5 the scores on a collegiate mathematics readiness assessment are approximately normally distributed with a mean of 680 and a standard deviation of 120. determine the percentage of scores between 690 and 900, to the nearest percent.
6 the heights of the members of a ski club are normally distributed. the average height is 64.7 inches with a standard deviation of 4.3 inches. determine the percentage of club members, to the nearest percent, who are between 67 inches and 72 inches tall.
7 a population is normally distributed with a mean of 23 and a standard deviation of 1.2. the percentage of the population that falls below 21, to the nearest hundredth, is
- 0.05
- 8.29
- 4.78
- 91.30
8 the distribution of the diameters of ball bearings made under a given manufacturing process is normally distributed with a mean of 4 cm and a standard deviation of 0.2 cm. what proportion of the ball bearings will have a diameter less than 3.7 cm?
- 0.0668
- 0.8664
- 0.4332
- 0.9500
9 the mean intelligence quotient (iq) score is 100, with a standard deviation of 15, and the scores are normally distributed. given this information, the approximate percentage of the population with an iq greater than 130 is closest to
- 2%
- 48%
- 31%
- 95%
Problem 1
Step1: Recall variability and standard deviation
Standard deviation ($\sigma$) measures variability. Larger $\sigma$ means more variability.
Step2: Compare $\sigma_1$ and $\sigma_2$
Given $\sigma_2 > \sigma_1$, so data set with $\sigma_2$ (data set two) has greater variability.
Step1: Recall 95% rule for normal distribution
For normal distribution, about 95% of data is within $\mu \pm 2\sigma$.
Step2: Set up equation
Mean $\mu = 12$, range is 8 to 16. So $12 - 2\sigma = 8$ and $12 + 2\sigma = 16$. Solve $12 - 2\sigma = 8$: $2\sigma = 12 - 8 = 4$, so $\sigma = 2$.
Step1: Calculate z - scores
Mean $\mu = 64$, $\sigma = 2.75$. For $x = 69.5$, $z=\frac{69.5 - 64}{2.75}=\frac{5.5}{2.75}=2$.
Step2: Use normal distribution properties
We want $P(64 < X < 69.5)=P(0 < Z < 2)$ (since $\mu = 64$ is the mean, $z = 0$ at $\mu$). From normal table, $P(Z < 2)=0.9772$, $P(Z < 0)=0.5$. So $0.9772 - 0.5 = 0.4772\approx 48\%$.
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- Data set two has the greater variability.