
Concept explainers
(a)
Generate the 1,000 random numbers between 1 through 5.
Find the expected
(a)

Answer to Problem 64CE
The mean value for the numbers between 1 and 5 is 3.
The standard deviation value for the numbers between 1 and 5 is 1.414.
Explanation of Solution
Calculation:
Generate 1,000 random numbers:
The formula to generate random numbers between two numbers by using EXCEL is RANDBETWEEN (bottom, top).
Software procedure:
Step by step procedure to obtain the 1,000 random numbers by using EXCEL software:
- • Open EXCEL.
- • In cell A1, enter the value as “=RANDBETWEEN(1,5)”.
Output using EXCEL software is given as follows:
4 | 5 | 2 | 4 | 5 | 2 | 2 | 2 | 4 | 5 | ... | 5 | 5 | 3 | 5 | 1 | 2 | 4 | 4 | 2 | 3 |
3 | 2 | 4 | 5 | 5 | 3 | 3 | 4 | 5 | 3 | … | 5 | 5 | 4 | 5 | 2 | 4 | 1 | 5 | 5 | 2 |
3 | 4 | 5 | 2 | 5 | 5 | 2 | 4 | 2 | 1 | … | 1 | 2 | 4 | 2 | 4 | 3 | 1 | 5 | 3 | 5 |
5 | 4 | 5 | 4 | 5 | 3 | 2 | 1 | 2 | 5 | … | 1 | 3 | 3 | 1 | 5 | 4 | 3 | 2 | 2 | 5 |
5 | 4 | 2 | 1 | 3 | 2 | 2 | 3 | 5 | 4 | … | 4 | 2 | 5 | 4 | 3 | 2 | 4 | 4 | 4 | 3 |
4 | 4 | 3 | 2 | 4 | 3 | 5 | 3 | 4 | 5 | … | 2 | 3 | 4 | 2 | 5 | 1 | 4 | 1 | 5 | 3 |
2 | 5 | 4 | 5 | 2 | 4 | 5 | 4 | 3 | 4 | … | 3 | 5 | 3 | 2 | 2 | 5 | 1 | 5 | 5 | 4 |
4 | 5 | 1 | 2 | 5 | 1 | 2 | 5 | 5 | 3 | … | 4 | 1 | 3 | 4 | 4 | 3 | 5 | 1 | 3 | 3 |
4 | 4 | 1 | 1 | 3 | 5 | 4 | 4 | 2 | 2 | … | 1 | 3 | 1 | 1 | 1 | 3 | 3 | 5 | 1 | 5 |
2 | 1 | 1 | 4 | 1 | 1 | 3 | 4 | 1 | 2 | … | 5 | 5 | 3 | 5 | 1 | 2 | 4 | 4 | 2 | 3 |
Uniform distribution:
The random variable with a finite number of integers from a and b follows uniform distribution with parameters a and b.
The
The formula for mean is,
The formula for standard deviation is,
The mean value for values between 1 and 5 is,
Thus, the value of mean for 1,000 random numbers between 1 and 5 is 3.
The standard deviation value for values between 1 and 5 is,
Thus, the value of standard deviation value for 1,000 random numbers between 1 and 5 is 1.414.
(b)
Find the sample mean and standard deviation.
(b)

Answer to Problem 64CE
The sample mean value for the numbers between 1 and 5 is 3.013.
The sample standard deviation value for the numbers between 1 and 5 is 1.4137.
Explanation of Solution
Calculation:
Answer will vary. One of the possible answer is given below:
Sample mean:
Software procedure:
Step by step procedure to obtain the sample mean for 1,000 random numbers by using EXCEL software:
- • Enter the formula “=AVERAGE (A1:A1000).
Output using EXCEL software is given as follows:
Thus, the value of sample mean for 1,000 random numbers between 1 and 5 is 3.013.
Sample standard deviation:
Software procedure:
Step by step procedure to obtain the sample mean for 1,000 random numbers by using EXCEL software:
- • Enter the formula “=STDEVA (A1:A1000).
Output using EXCEL software is given as follows:
Thus, the value of sample standard deviation for 1,000 random numbers between 1 and 5 is 1.4137.
(c)
Check the sample is consistent with uniform model.
(c)

Answer to Problem 64CE
The sample is consistent with uniform model.
Explanation of Solution
Calculation:
Answer will vary. One of the possible answer is given:
From the result of part (a): the results are drawn from the uniform distribution.
The mean value for the numbers between 1 and 5 is 3 and the standard deviation value for the numbers between 1 and 5 is 1.414.
From the result of part (b): the results are drawn from the samples.
The sample mean value for the numbers between 1 and 5 is 3.013 and the sample standard deviation value for the numbers between 1 and 5 is 1.4137.
By observing the results, in both cases the results are same. Thus, the sample is consistent with uniform model.
(d)
Give the EXCEL formulae used to generate random numbers.
(d)

Explanation of Solution
Generate the 1,000 random numbers between 1 through 5 is “=RANDBETWEEN(1,5)”.
Want to see more full solutions like this?
Chapter 6 Solutions
Applied Statistics in Business and Economics
- 1. [20] The joint PDF of RVs X and Y is given by xe-(z+y), r>0, y > 0, fx,y(x, y) = 0, otherwise. (a) Find P(0X≤1, 1arrow_forward4. [20] Let {X1,..., X} be a random sample from a continuous distribution with PDF f(x; 0) = { Axe 5 0, x > 0, otherwise. where > 0 is an unknown parameter. Let {x1,...,xn} be an observed sample. (a) Find the value of c in the PDF. (b) Find the likelihood function of 0. (c) Find the MLE, Ô, of 0. (d) Find the bias and MSE of 0.arrow_forward3. [20] Let {X1,..., Xn} be a random sample from a binomial distribution Bin(30, p), where p (0, 1) is unknown. Let {x1,...,xn} be an observed sample. (a) Find the likelihood function of p. (b) Find the MLE, p, of p. (c) Find the bias and MSE of p.arrow_forwardGiven the sample space: ΩΞ = {a,b,c,d,e,f} and events: {a,b,e,f} A = {a, b, c, d}, B = {c, d, e, f}, and C = {a, b, e, f} For parts a-c: determine the outcomes in each of the provided sets. Use proper set notation. a. (ACB) C (AN (BUC) C) U (AN (BUC)) AC UBC UCC b. C. d. If the outcomes in 2 are equally likely, calculate P(AN BNC).arrow_forwardSuppose a sample of O-rings was obtained and the wall thickness (in inches) of each was recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. Click here to view the table of critical values for normal probability plots. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. 0.191 0.186 0.201 0.2005 0.203 0.210 0.234 0.248 0.260 0.273 0.281 0.290 0.305 0.310 0.308 0.311 Using the correlation coefficient of the normal probability plot, is it reasonable to conclude that the population is normally distributed? Select the correct choice below and fill in the answer boxes within your choice. (Round to three decimal places as needed.) ○ A. Yes. The correlation between the expected z-scores and the observed data, , exceeds the critical value, . Therefore, it is reasonable to conclude that the data come from a normal population. ○…arrow_forwardding question ypothesis at a=0.01 and at a = 37. Consider the following hypotheses: 20 Ho: μ=12 HA: μ12 Find the p-value for this hypothesis test based on the following sample information. a. x=11; s= 3.2; n = 36 b. x = 13; s=3.2; n = 36 C. c. d. x = 11; s= 2.8; n=36 x = 11; s= 2.8; n = 49arrow_forward13. A pharmaceutical company has developed a new drug for depression. There is a concern, however, that the drug also raises the blood pressure of its users. A researcher wants to conduct a test to validate this claim. Would the manager of the pharmaceutical company be more concerned about a Type I error or a Type II error? Explain.arrow_forwardFind the z score that corresponds to the given area 30% below z.arrow_forwardFind the following probability P(z<-.24)arrow_forward3. Explain why the following statements are not correct. a. "With my methodological approach, I can reduce the Type I error with the given sample information without changing the Type II error." b. "I have already decided how much of the Type I error I am going to allow. A bigger sample will not change either the Type I or Type II error." C. "I can reduce the Type II error by making it difficult to reject the null hypothesis." d. "By making it easy to reject the null hypothesis, I am reducing the Type I error."arrow_forwardGiven the following sample data values: 7, 12, 15, 9, 15, 13, 12, 10, 18,12 Find the following: a) Σ x= b) x² = c) x = n d) Median = e) Midrange x = (Enter a whole number) (Enter a whole number) (use one decimal place accuracy) (use one decimal place accuracy) (use one decimal place accuracy) f) the range= g) the variance, s² (Enter a whole number) f) Standard Deviation, s = (use one decimal place accuracy) Use the formula s² ·Σx² -(x)² n(n-1) nΣ x²-(x)² 2 Use the formula s = n(n-1) (use one decimal place accuracy)arrow_forwardTable of hours of television watched per week: 11 15 24 34 36 22 20 30 12 32 24 36 42 36 42 26 37 39 48 35 26 29 27 81276 40 54 47 KARKE 31 35 42 75 35 46 36 42 65 28 54 65 28 23 28 23669 34 43 35 36 16 19 19 28212 Using the data above, construct a frequency table according the following classes: Number of Hours Frequency Relative Frequency 10-19 20-29 |30-39 40-49 50-59 60-69 70-79 80-89 From the frequency table above, find a) the lower class limits b) the upper class limits c) the class width d) the class boundaries Statistics 300 Frequency Tables and Pictures of Data, page 2 Using your frequency table, construct a frequency and a relative frequency histogram labeling both axes.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman





