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Chapter 6: Problem 47

Determine each of the following areas under the standard normal (z) curve: a. To the left of -1.28 b. To the right of 1.28 c. Between -1 and 2 d. To the right of 0 e. To the right of -5 f. Between -1.6 and 2.5 g. To the left of 0.23

Short Answer

Expert verified
a. 0.1003, b. 0.1003, c. 0.8186, d. 0.5000, e. approx. 1, f. 0.9452, g. 0.5910.

Step by step solution

01

Understanding the z-score

A z-score measures how many standard deviations an element, X, is from the mean of the distribution, in this case, 0. Negative z-scores are below the mean, positive ones are above it. To interpret z-scores (which will be the values given in the exercise), one can use a Z-table or online calculator.
02

Calculate area under the curve

(a) To the left of -1.28, we look up this z-score in the z-table or use an online calculator. (b) To the right of 1.28, we calculate 1 minus the value for 1.28 from the z-table or online calculator. (c) Between -1 and 2, we subtract the z-score for -1 from the z-score for 2. (d) To the right of 0 is simply 0.5, as zero is exactly in the middle. (e) To the right of -5 is almost 1 as -5 is far away to the left under the curve. (f) Between -1.6 and 2.5, we subtract the z-score for -1.6 from that for 2.5. (g) To the left of 0.23, we simply use the value from the table or online calculator.
03

Use of Z-table or Online Calculator

For each part (a-g), we use a Z-table or online calculator to find the appropriate values. Students need to search the z-score and find the corresponding area under the curve.

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Most popular questions from this chapter

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1 .\) Determine each of the following probabilities: a. \(P(z < 0.10)\) b. \(P(z < -0.10)\) c. \(P(0.40 < z < 0.85)\) d. \(P(-0.85 < z < -0.40)\) e. \(P(-0.40 < z < 0.85)\) f. \(P(z > \- 1.25)\) g. \(P(z < -1.50\) or \(z > 2.50)\)

A pizza company advertises that it puts 0.5 pound of real mozzarella cheese on its medium-sized pizzas. In fact, the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pound and a standard deviation of 0.025 pound. a. What is the probability that the amount of cheese on a medium pizza is between 0.525 and 0.550 pound? b. What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? c. What is the probability that three randomly selected medium pizzas each have at least 0.475 pound of cheese?

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

A symptom validity test (SVT) is sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for Posttraumatic Stress Disorder: Application of the Binomial Distribution" ( Journal of Anxiety Disorders [2008]: 1297-1302) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.70\) to calculate various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: i. \(\quad P(x=42)\) ii. \(P(x<42)\) iii. \(P(x \leq 42)\) c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96\), they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the random variable \(x\) as \(x=\) the number of people who actually show up for a sold-out flight on this plane
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