Explanation Given: Mean μ = 60 and standard deviation σ = 10 . Approach: The expression to obtain the probabilities are as follows, P ( X > a ) = P ( Z > a − μ σ ) … ( 1 ) P ( X < a ) = P ( Z < a − μ σ ) … ( 2 ) Here, P ( X ) is the Probability of an event, Z is the range, μ is the mean, σ is a standard deviation

12th Edition

ISBN: 8220103649001

Author: Tan

Publisher: CENGAGE L

See similar textbooks

Concept explainers

Continuous Probability Distributions

Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could co…

Normal Distribution

Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a…

Videos

Question

Chapter 8.6, Problem 8E

To determine

The percentage of applicants passed the written qualifying examination.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.6, Problem 7E

Chapter 8.6, Problem 9E

Students have asked these similar questions

1. Find the solution set of In(x) sin(x) ≤ 0, for x = [0,14].

Suppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a. Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.) I worked out the Upper Limit, but I can't seem to arrive at the correct answer for the Lower Limit. What is the Lower Limit?…

4. Consider Chebychev's equation (1 - x²)y" - xy + λy = 0 with boundary conditions y(-1) = 0 and y(1) = 0, where X is a constant. (a) Show that Chebychev's equation can be expressed in Sturm-Liouville form d · (py') + qy + Ary = 0, dx y(1) = 0, y(-1) = 0, where p(x) = (1 = x²) 1/2, q(x) = 0 and r(x) = (1 − x²)-1/2 (b) Show that the eigenfunctions of the Sturm-Liouville equation are extremals of the functional A[y], where A[y] = I[y] J[y]' and I[y] and [y] are defined by - I [y] = √, (my² — qy²) dx and J[y] = [[", ry² dx. Explain briefly how to use this to obtain estimates of the smallest eigenvalue >1. 1 (c) Let k > be a parameter. Explain why the functions y(x) = (1-x²) are suitable 4 trial functions for estimating the smallest eigenvalue. Show that the value of A[y] for these trial functions is 4k2 A[y] = = 4k - 1' and use this to estimate the smallest eigenvalue \1. Hint: L₁ x²(1 − ²)³¹ dr = 1 (1 - x²)³ dx (ẞ > 0). 2ẞ

Chapter 8 Solutions

EBK FINITE MATHEMATICS FOR THE MANAGERI

Ch. 8.1 - What is a random variable? Give an example.Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Three balls are selected at random without...Ch. 8.1 - 2.A coin is tossed four times. Let the random...Ch. 8.1 - A die is rolled repeatedly until a 6 falls...Ch. 8.1 - Cards are selected one at a time without...Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The percentage of applicants passed the written qualifying examination.

Solution Summary: The author explains how the percentage of applicants passed the written qualifying examination is 15.87%.

Concept explainers

Videos

The percentage of applicants passed the written qualifying examination.

Want to see the full answer?

Chapter 8 Solutions