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.)
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, z, x̅, or p̂) 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.
Suppose you have a skewed distribution with mean $48,000 and standard deviation $2,000. You take a sample of size 64 and compute the mean of the sample. That x̅ is one sample mean out of the distribution of all possible sample means. Use this information to answer questions 1–3.
Answer: Approximately normal because the variable is numeric and n ≥ 30
Answer: μx̅ = $48,000
Answer: σx̅ = σ/√n = $2000/√64 ⇒ σx̅ = $250
Remark: The SEM (standard error of the mean) is an important concept in statistics, as we’ll see further in Chapter 9. Make sure you understand the concept and the calculation.
Solution: (a) Given: μ = 800, σ = 50, n = 100
P(x̅ ≤ 780) = normalcdf(−10^99, 780, 800, 50/√(100)) = 3.1686E-5 ≈ 0.00003
(You can also give the probability as <0.0001.)
Common mistake: Do not give the probability as 3.1686. Probabilities are never greater than 1.
(b) If the manufacturer’s claim is true, there are only three chances in a hundred thousand of getting a sample mean this low. It’s very unlikely that the manufacturer’s claim is true.
Given: μ = 5.00, σ = 0.05, n = 15.
If 15 bags weigh ∑x = 75.6 lb, the sample mean is x̅ = ∑x/n = 75.6/15 = 5.04 lb. A sample weighing 75.6 lb total will have a sample mean of 5.04 lb, so this is really just another problem in finding the probability of turning up a sample mean in a given range.
P(∑x > 75.6) = P(x̅ > 5.04) = normalcdf(75.6/15, 10^99, 5.00, 0.05/√(15)) = 9.7295E-4 ≈ 0.0010
This is just under one chance in a thousand.
Solution: (a) Remember that “describe the distribution” means shape, center and spread. You can always get center and spread, but if the test for normal approximation fails then you can’t say anything about the shape.
Answer: normally distributed with mean = 0.72, s.d. (standard error) = 0.02
Common mistake: Don’t write n≥30 when testing the normal approximation. Though n is indeed greater than 30 in this problem, it’s irrelevant because the n≥30 test applies to numeric data, but in this problem you have binomial data.
(b) If you
stored the computed SEP
in part (a), then
P(0.70 ≤ p̂ ≤ 0.74) =
normalcdf(0.70, 0.74, 0.72, A) =
Otherwise, normalcdf(0.70, 0.74, 0.72, √(.72*(1−.72)/500))=.6808
Remark: Always check for reasonableness. 70% and 74% are one standard error below and above the mean, so you know from the Empirical Rule that about 68% of the data should be within that region.