MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- 5. For a certain disease, 40% of untreated patients can be expected to improve within a week. We observe a population of 80 untreated patients and record the percentage who improve within one week. According to the Central Limit Theorem, such percentages are approximately normally distributed. Find the mean and standard deviation of this normal distribution. Round the standard deviation to two decimal places.arrow_forwardSuppose that the average salary of health systems professionals is normally distributed with µ=KES 40,000 and σ = KES 10,000. (20 Marks) What proportion of professionals will earn KES 24,800 or less? What proportion of professionals will earn KES 53,500 or more? What proportion of professionals will earn between KES 45,000 and KES 57,000? Calculate the 80th percentile. Calculate the 27th percentilearrow_forwardanswer B C D Earrow_forward
- 4. A distribution of scores has = 70. The z-score for X = 65 is computed and a value of z = +1.00. Regardless of the value of the standard deviation, why must this z-score be incorrect?arrow_forwardThe weight in kilograms of the citizens of Fredonia has a mean of 50 kg and standard deviation of 24 kg (a) Find the mean of the sampling distribution of the sample mean. (b) Find the standard deviation of the sampling distribution of the sample mean.arrow_forward4. A normal population has the mean of 20 and the variance of 100. A random sample of size n = 69 is selected. (a) Find the standard deviation of the sample mean. (b) How large must the sample be if you want to halve the standard deviation of the sample mean?arrow_forward
- 5. On an examination the average grade was 74 and the standard deviation was 7. If 12% of the class are given A's, and the grades are curved to follow normal distribution, what is the lowest possible A and the highest possible B? P(Z < = 1-0.12 = 0.88 1o nou X1 - 74 X1 = 7arrow_forwardIn a population of a certain species of newt, the mean length of the newts is μ= 63 inches and the standard deviation is σ= 3.4 inches. You collect a random sample of n=16 newts. The bell curve below represents the distribution of these sample means. The scale on the horizontal axis is the standard error of the sampling distribution. Complete the indicated boxes, correct to two decimal places. mean of distribution= standard error= Note: The left box is 2 standard errors below the mean. The middle box is the mean. The right box is 2 standard errors above the mean.arrow_forwardThis Is A REVIEW! Please help and show work on paperarrow_forward
- ou should complete the table as part of the procedure of finding the variance and standard deviation. Xi (Xi-(mu)) | (Xi-(mu)) | 49 53 60 60 62 63 64 69 73 87 [Note, this is the same set of data points used in problem 6. How does this compare to the standard deviation computed in problem 6? Why might this be smaller?]arrow_forward5. The distribution of heights of a certain breed of terrier has a mean of 72 centimeters and a standard deviation of 10 centimeters, whereas the distribution of heights of a certain breed of poodle has a mcan of 28 centimeters with a standard deviation of 5 centimcters. Assume the distributions of both types of breeds are normally distributed. (n) Suppose we are taking a random sample of 64 terriers. What is the sampling distri- bution of mean heights of the breed of terriers? (b) Find the probability that the sample mean for a random sample of heights of 61 terriers excoedls 75 centimetcrs. (c) Find the probability that the sample mean for a random sample of heights of 64 terriers excoeds the sample mean for a random sample of heights of 100 poodles by at lenst 44.2 centimeters.arrow_forward
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