TC3 → Stan Brown → Statistics → Chapter 7 Practice Quiz
revised 31 Mar 2013

Practice Quiz with Solutions: Chapter 7 (20 min)

Before looking at these solutions, please work the practice quiz.

Disclaimer: This quiz is representative of the level of difficulty you should expect, but it doesn’t include every single topic from the week’s work. The real quiz may include some other topics and may skip some that are in this practice quiz. (The real quiz also may not word questions in the same way as the practice quiz. You should focus on the concepts, not a particular form of words.)

See also: How to Take a Math Test
Normal Calculations on TI-83/84 or TI-89

These solutions show about the same level of work I expect from you, though I often add some extra commentary. Please see Show Your Work for the what, why, and how.

Caution: Your TI-83 shows the curve and shaded area correctly, but because of the limitations of screen area the TI-83 graphs are not labeled correctly. Your lower bound (x or z) must always be under the left edge of the shaded area, and your upper bound must always be under the right edge of the shaded area. The area number must not be below the axis at all, but above and next to the shaded area.

1(points: 3½) Suppose that the new Ford Behemoth SUV gets a mean of shaded normal curve for this problem 6.20 miles to the gallon, normally distributed with standard deviation 0.90 mpg. Jane, who has taken a statistics class, found that 85% of Ford Behemoths get better gas mileage than hers does. What is her car’s gas mileage, to two decimal places?

Solution: You should have a normal curve with μ = 6.2 and σ = 0.9. The right-hand 85% of the curve should be shaded in.

invNorm(1–.85, 6.2, 0.9) = 5.2672 → 5.27 mpg

Common mistake: Some students calculate invNorm(0.85,...) instead of invNorm(1−.85,...). Remember that invNorm wants area to left, but the 85% is area to right.

To guard against mistakes like this, always ask yourself whether your answer is reasonable. Here you know that 85% of cars get more mileage than hers. If 50% of cars got better mileage than hers, her car would be average (6.2 mpg). But since more than 50% of cars get better mileage than hers, her car’s mileage must be below average, less than 6.2. An answer greater than 6.2 should ring your alarm bells.

 
2(points: 1)  What is the percentile rank of Jane’s car?

Solution: If 85% of cars get better mileage than hers, then 15% get worse mileage. Her car is at the 15th %ile.

3(points: 3½) The times to assemble a particular product part are shaded normal curve for assembly workers normally distributed with a mean of 47.5 minutes and a standard deviation of 8.5 minutes. What percent of assembly workers require more than 60 minutes?

Solution: The problem asks for P(x>60). You should have a normal curve with μ = 47.5 and σ = 8.5 as given above. 60 minutes is between one and two s.d. above the mean; therefore the shaded tail will be of moderate size. You could do ShadeNorm (as shown at right) and read off the answer, 0.0707 or 7.07%. Or you could use normalcdf:

P(x>60) = normalcdf(60, 10^99, 47.5, 8.5) = .0707 = 7.07%

4(points: 1½) The standard normal curve is plotted in terms of _____ (write in “z” or “x”). It has a mean of _____ and a standard deviation of ___.

Answer: The standard normal curve is plotted in terms of z. It has a mean of 0 and a standard deviation of 1.

5(points: 3) The lifetimes of Everlast AA batteries are normally distributed. Between x=28.2 and x=34.1 hours, the area under the curve is 0.0742. Give two interpretations of this fact, in English.

Answer:

Remember: probability of one = proportion of all. Sometimes one interpretation is more natural, sometimes the other.

6(points: 2½)  Fourteen rats were sent into space, and on their return to Earth their red-blood-cell masses were measured. Here are the data, in milliliters:

 8.59   6.87   7.00   6.39   7.43   9.79   9.30   8.64   7.89   8.80   7.54   7.21   6.85   8.03 

Are those figures normally distributed? Support your answer, but you need not draw a graph.

normal prrobability plot for the rats

Solution: Use MATH200A part 4, the normality check. You’ll see a duplicate of the screen shot at right. The points are very nearly a straight line, confirmed by r=.9838 and CRIT=.9351. r>CRIT, and therefore you can say that the rats’ red-blood-cell masses are normally distributed.


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