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MATH 110 All Exams Statistics Questions with Complete
Solutions 2025- Portage Learning
1.Define each of the following: a) Observation b) Element c) Variable Observation- all the information collected for each element in a study Element- in a data set, the individual and unique entry about which data has been collected, analyzed and presented in the same manner Variable- a particular, measurable attribute that the researcher believes is needed to describe the element in their study. 2.Explain outliers An outlier is a value which is out of place compared to the other values. It may be too large or too small compared to the other values 3.Look at the following data and see if you can identify any outliers: 53 786 789 821 794 805 63 777 814 2333 783 811 795 788 780 Outliers: 53 63 2333
a) How many were burgers? b) How many were fish? a) Burgers, 2900(0.12)= b) Fish, 2900(0.28)= Mod 2 Exam 1.During an hour at a fast food restaurant, the following types of sandwiches are ordered: Turkey Turkey Cheeseburger Hamburger Fish Chicken Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Chicken Fish Turkey Fish Hamburger Fish Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken Turkey Fish Hamburger Chicken Fish a) Make a frequency distribution for this data. b) Make a relative frequency distribution for this data. Include relative percentages on this table. a.
- Consider the following data: {29, 20, 24, 18, 32, 21} a) Find the sample mean of this data. b) Find the range of this data. c) Find the sample standard deviation of this data. d) Find the coefficient of variation. a.
- Suppose that you have a set of data that has a mean of 49 and a standard deviation of 8. a) Is the point 57 above, below, or the same as the mean. How many standard deviations is 57 from the mean. b) Is the point 33 above, below, or the same as the mean. How many standard deviations is 33 from the mean. c) Is the point 31 above, below, or the same as the mean. How many standard deviations is 31 from the mean. d) Is the point 79 above, below, or the same as the mean. How many standard deviations is 79 from the mean. a) The data point 57 is above the mean. Now use the z-score to determine how many standard deviations 57 is above the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by: The z-score is 1, so the data point 57 is 1 standard deviation above the mean. b) The data point 33 is below the mean. Now use the z-score to determine how many standard deviations 33 is below the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by: The z-score is -2, so the data point 33 is 2 standard deviations below the mean (the negative sign indicates that the point is below the mean). c) The data point 31 is below the mean. Now use the z-score to determine how many standard deviations 31 is below the mean. We are told that the mean is 49 and the standard deviation is 8. So, the z-score is given by:
- Suppose you are going to make a password that consists of 5 characters chosen from {1,2,4,9,d,i,k,m,n,w,z}. How many different passwords can you make if you cannot use any character more than once in each password?
- Suppose A and B are two events with probabilities: P(A)=.35, P(Bc^ )=.45, P(A∩B)=.25. Find the following: a) P(A B). b) P(Ac^ ). c) P(B). a. For P(A B). Use P(A B)=P(A)+P(B)-P(A∩B). But for this equation, we need P(B) which we can find by using P(B)=1-P(B c ). So, P(B)=1-.45= .55. P(A B)=.35+.55-.25=. b. For P(A c ). Use P(A)=1-P(A c ) which may be rearranged to (A c )=1-P(A). P(A c )=1-.35=.65. c. For P(B). Use (B)=1-P(Bc^ ). P(B)=1-.45=.55.
- Suppose A and B are two events with probabilities: P(A c )=.50, P(B)=.65, P(A∩B)=.30. a) What is (A│B)? b) What is (B│A)?
- In a manufacturing plant, three machines A, B, and C produce 30 %, 20 %, and 50 %, respectively, of the total parts production. The company's quality control department determined that 3 % of the parts produced by machine A, 2.5 % of the parts produced by machine B, and 4 % of the parts produced by machine C are defective. If a part is selected at random and found to be defective, what is the probability that it was produced by machine B?
- The probability that a certain type of battery in a smoke alarm will last 3 years or more is .85. The probability that a battery will last 7 years or more is .25. Suppose that the battery is 3 years old and is still working, what is the probability that the battery will last at least 7 years?
- Find each of the following probabilities: (use standard normal distribution table to get z- score) a. Find P(Z ≤ 1.27). b. Find P(Z ≥ -.73). c. Find P(-.09 ≤ Z ≤ .86). a. P(Z ≤ 1.27) =0. b. P(Z ≥ -0.73= 1- P(Z ≤ -0.73)=1- 0.23270=0. c. P(-0.09 ≤ Z ≤0 .86)= P(Z≤0 .86)- P(Z≤ -0.09) 0.80511-0.46414=.
- A company manufactures a large number of rods. The lengths of the rods are normally distributed with a mean length of 4.0 inches and a standard deviation of .75 inches. If you choose a rod at random, what is the probability that the rod you chose will be: a) Less than 3.0 inches? b) Greater than 3.7 inches? c) Between 3.5 inches and 4.3 inches? a. We must find the z-score for x=3.0:
So, we want P(Z ≤ -1.33). From the table, we find. P(Z ≤ -1.33)=.09176. b. We must find the z-score for x=3.7: So, we want P(Z ≥-.4). Since this is greater than, we must use: P(Z ≥-.4)=1.0-P(Z ≤-.4)=1- .34458= .65542. c. We must find the z-score for x=3.5: and the z-score for x=4.3: So, we want P(-.67 ≤ Z ≤ .4) P(-.67 ≤ Z ≤ .4)=P(Z≤ .4)-P( Z ≤ -.67). P(-.67 ≤ Z ≤ .4)=.65542-.25143=.40399.
- A life insurance salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as follows: Policies Sold Probability, f(x) Per Day
Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data.
Calculate the z-score: So, we want to find P(Z < .782) on the standard normal probability distribution table. Recall that P(Z < .78) = .78230. Therefore, there is a 0.78230 probability that a simple random sample of 30 nurses will have a mean time of 155 seconds or less.
- Suppose that in a very large city 9.8 % of the people have more than two jobs. Suppose that you take a random sample of 70 people in that city, what is the probability that 9 % or more of the 70 have more than two jobs? Now we find the z-score: We want P(Z>-0.23). From the standard normal table, we find: P(Z>-.23)=1- P(Z<-.23)=1-.40905=.59095. So there is a .59095 probability that the percentage of the sample that have more than two jobs is more than 9 %. Exam 6
- A new drug for migraine headaches has been introduced to the market. You would like to know if migraine patients prefer the old drug or the new drug. So, you sample 190 patients that have used both drugs and you find that 52% of the sampled patients prefer the new drug, 48% of the patients prefer the old drug. Find the 90% confidence limit for the proportion of all patients that prefer the new drug. Can you be 90% confident that a majority of all the patients prefer the new drug?
p=.52 n = 190 Based on a confidence limit of 90 %, we find in table 6.1 that z=1. So, the 90% confidence limit is: Notice that the proportion that like the new drug may be as small as .46 which is less than the .5. So, we cannot be 90% confident that a majority of all the patients will prefer the new drug.
- Suppose that you are a nurse and you are assigned to do checkups of people one day per week in a certain village. You have a total of 300 patients in the village. You have the option of doing the checkups in the mornings or in the afternoons. Therefore, you ask 35 patients and find that 62% prefer afternoon appointments while 38% prefer morning appointments. Find the 95% confidence limit for the proportion of all patients that prefer afternoon appointments. Since 62% prefer afternoon, we set P = .62. As we mentioned previously, we estimate p by P. So, p=.62. The total population is 300, so set N=300. A total of 35 patients were surveyed, so Based on a confidence limit of 95 %, we find in table 6.1 that z=1.96. Now, we can substitute all of these values into our equation: So the proportion of the total who prefer afternoon appointments is between .469 and .771. Exam 7
- a) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
This is a left-tailed test, so we must find a z that satisfies P(Z<z)=.02. In the standard normal table, we find z. 02 = -2.05. For a left-tailed test, we will reject the null hypothesis if the z-score is less than -2.05. We find the z-score: Notice that since the z-score is less than -2.05, we reject the null hypothesis.
- A mayor claims that the unemployment rate in his city is 4 %. Many people think that the unemployment rate is higher. So, 95 residents of the city are contacted and it is found that 8 of them are unemployed. Can the mayor’s claim be supported to a level of significance of α = .02, test the hypothesis. H 0 : p=.04. H 1 : p>.04. Since this is a right-tailed test. z=1-0.02=0.98=2.05. Since this is a right-tailed test, and the z-score is greater than 2.05, we reject the null hypothesis. Exam 8 Suppose we have independent random samples of size n 1 = 420 and n 2 = 510. The proportions of success in the two samples are p 1 = .38 and p 2 = .43. Find the 99% confidence interval for the difference in the two population proportions. Answer the following questions:
- Multiple choice: Which equation would you use to solve this problem?
A. B. C. D.
- List the values you would insert into that equation.
- State the final answer to the problem From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2: B. So, the interval is (.-0.1333, -0.03326).
b) Since the entire confidence interval is positive, we can be 90 % sure that there is a difference in the means of the two populations. A head librarian supervises a number of libraries in a large county. He wants to know if full- time library workers and part-time library workers re-shelve books at the same rate. So, he checks the records of 40 full-time library workers and finds that they re-shelve an average of 185 books per hour with a standard deviation of 17.1 books per hour. The records of 40 part- time library show that they re-shelve an average of 190 books per hour with a standard deviation of 9.2 books per hour. Using a level of significance of α=.10, is there enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part-time workers? Answer the following questions:
- Multiple choice: Which equation would you use to solve this problem? A. B. C. D.
- List the values you would insert into that equation.
- State the final answer to the problem H 0 : μ 1 - μ 2 = 0 H 1 : μ 1 - μ 2 ≠0. Since this is a two-tailed test, we must find the z that satisfies: P(Z<z)=.1/2=.05 and P(Z > z)=.1/2=.05. In the standard normal table, z=-1.645 and z=1.645. We will reject the null hypothesis if the z-score is less than -1.645 or the z-score is greater than 1.645. We now find the z-score:
Since the z-score is between -1.645 and 1.645, we do not reject the null hypothesis. Consider the following dependent random samples Observations 1 2 3 4 5 6 x-values 8.1 7.6 8.3 8.4 7.9 7.0 y- values 8.4 8.4 8.5 8.9 8.1 7. a) Determine the difference between each set of points, xi - yi b) Do hypothesis testing to see if μd < 0 at the α = .025. Since we are testing whether or not μd < 0, then our null and alternate hypothesis will be set as follows: H 0 : μd = 0 H 1 : μd < 0 n=6. This is a left-tailed test. Note that for t. 025 = -2.571 for 6- 1 = 5. We find the mean in the usual way: The sample standard deviation is given by: Then using the mean, d = -.4333, and the standard deviation, sd= .2422, that we found above: Since t < t. 025 , we reject the null hypothesis.
- A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and without the drink are given below: Runner 1 2 3 4 5 6 7 8 x-time (before) 254 276 276 265 271 273 268 281 y-time (after) 265 269 277 279 266 273 275 279 Find the 95 % confidence interval for mean of the differences, μd.
