Question

1. the environmental protection agency (epa) requires that the exhaust from each model of motor vehicle be tested for the level of several pollutants. the level of oxides of nitrogen (nox) in the exhaust of one light truck model was found to vary among individual trucks according to an approximately normal distribution with mean μ = 1.50 grams per mile driven and standard deviation σ = 0.25 gram per mile. which of the following best estimates the percent of light trucks of this model with nox levels greater than 2 grams per mile?

a. 2.5%
b. 5%
c. 16%
d. 32%

1. the figure shows a density curve that models the distribution of a quantitative variable. seven values are marked on the density curve. which of the following statements is true?

a. the mean of the distribution is e.
b. the median of the distribution is c.
c. the third quartile of the distribution is d.
d. the area under the curve between a and g is 1.

1. in 1965, the mean price of a new car was $2650 and the standard deviation was $1000. in 2019, the mean was $36,700 and the standard deviation was $9000. if a ford mustang cost $2300 in 1965 and $34,000 in 2019, in which year was it more expensive relative to other cars?

a. 1965, because the standardized score (z - score) is greater than in 2019.
b. 1965, because the standard deviation is smaller.
c. 2019, because the standardized score (z - score) is greater than in 1965.
d. 2019, because $34,000 is greater than $2300.

1. the distribution of employee salary at a large company is skewed to the right, with mean $75,100 and standard deviation $36,554. because the company had a productive year, the ceo will give every employee 2% of their annual salary as a rear - end reward, along with a $500 christmas bonus. so an employee with a $60,000 salary would receive an adjusted salary of (60,000)(1.02)+500 = $61,700. find the adjusted mean and standard deviation.

a. adjusted mean = $77,102 and adjusted standard deviation = $37,285.08.
b. adjusted mean = $76,602 and adjusted standard deviation = $37,285.08.
c. adjusted mean = $75,600 and adjusted standard deviation = $36,554.
d. adjusted mean = $75,600 and adjusted standard deviation = $37,785.08.

Explanation:

Step1: Calculate z-score for 2 grams

$z = \frac{2 - 1.5}{0.25} = 2$

Step2: Use empirical rule for normal distribution

P(z > 2) ≈ 2.5%

(Note: Without the figure, this question cannot be answered accurately. Assuming the density curve is right-skewed, the mean is to the right of the median. If E is rightmost, a might be true; but this is speculative without the figure.)

Step1: Calculate 1965 z-score

$z_{1965} = \frac{2300 - 2650}{1000} = -0.35$

Step2: Calculate 2019 z-score

$z_{2019} = \frac{34000 - 36700}{9000} ≈ -0.3$

Step3: Compare z-scores

-0.3 > -0.35, so 2019 z-score is greater

Step1: Calculate adjusted mean

$\text{Adjusted mean} = 75100×1.02 + 500 = 76602$

Step2: Calculate adjusted standard deviation

$\text{Adjusted SD} = 36554×1.02 = 37285.08$ (additions don't affect SD)

Answer:

a. 2.5%