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MATH 110 Module 1-10 Exams plus FINAL EXAM
BUNDLE- Portage Learning
MATH 110 - Introduction to Statistics
COMBINATION OF ALL EXAMS ( 1 - 10)
Exam 1
Exam Page 1 Define each of the following: a) Element. An element is described as "the individual and unique entry in a data set about which data has been collected, analyzed and presented in a same manner to differentiate" (Module 1). b) Variable. A variable is defined as a "particular measurable attribute that the researcher believes is needed to describe the element in their study" (Module 1). c) Data. Data (or the plural of datumn) is defined as things (such as numerical information, people, geographical areas,etc.) about which information can be collected and then analyzed.
Answer Key
Define each of the following: a) Element. a) The element of a data set is simply the individual and unique entry in a data set about which data has been collected, analyzed and presented in the same manner. b) Variable. b) A variable is a particular, measurable attribute that the researcher believes is needed to describe the element in their study. c) Data. c) Data are things about which information can be collected and analyzed.
The following pie chart shows the percentages of total items sold in a month in a certain fast food restaurant. Exam Page 4 The following pie chart shows the percentages of total items sold in a month in a certain fast food restaurant. A total of 4900 fast food items were sold during the month. How many were burgers? How many were french fries? 4900(.32)= 32% or 1,568 burgers were sold during the month. 4900(.18)= 18% or 882 french fries were sold during the month.
Instructor Comments
Very good.
Answer Key
The outliers are: 3 246 277
Exam 2
Exam Page 1
During an hour at a fast food restaurant, the following types of sandwiches are ordered: Turkey Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Fish Chicken 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 Chicken Hamburger Chicken Fish Chicken a) Make a frequency distribution for this data. Types of Frequency Sandwiches Turkey 8 A total of 4900 fast food items were sold during the month. How many were burgers? How many were french fries? Burgers : 4900(.32) = 1568 French Fries : 4900(.18) = 882
n x=(29+20+24+18+32+21) = 144 = 24 6 6 b) Find the range of this data. {18,20,21,24,29,32} Range is 14 (32-18)= c) Find the sample standard deviation of this data. s^2 =∑(xi-x)^2 = (18-24)^2 + (20-24)^2 +(21-24)^2 +(24-24)^2 +(29-24)2+(32-24)^2 = 36+16+9+0+25+64= 150 = n- 1 6 - 1 5 5 s=√s^2 = √30 = 5. d) Find the coefficient of variation. cov=standard deviation100=5.477100 =22. mean 24 Exam Page 4 Suppose that you have a set of data that has a mean of 65 and a standard deviation of 10. a) Is the point 75 above, below, or the same as the mean. How many standard deviations is 75 from the mean. x z=x-u= 75 - 65 = o 10 z= The point 75 is above the mean (because it is a positive number), meaning that the data point is one standard deviation above the mean.
b) Is the point 85 above, below, or the same as the mean. How many standard deviations is 85 from the mean. x* z=x-u= 85 - 65 = o 10 z= The point 85 is above the mean (because it is a positive number), meaning that the data point is two standard deviations above the mean. c) Is the point 57.5 above, below, or the same as the mean. How many standard deviations is 57. from the mean. x* z=x-u= 57.5- 65 =-0. o 10 z=-0. The point 57.5 is below the mean (because it is a negative number), meaning that the data point is .75 standard deviations below the mean. d) Is the point 107 above, below, or the same as the mean. How many standard deviations is 107 from the mean. x* z=x-u= 107 - 65 =4. o 10 z=4. The point 107 is above the mean (because it is a positive number), meaning that the data point is 4.2 standard deviations above the mean. Exam Page 5 Consider the following set of data: {22, 14, 35, 49, 8, 18, 30, 44} a) Find the median. {8,14,18,22,30,35,44,49} Median=22+30=
Suppose you are going to make a password that consists of 4 characters chosen from {2,7,8,c,f,k,t,z}. How many different passwords can you make if you cannot use any character more than once in each password? n=8 (only 8 characters to choose from) r=4 (four character passwords P(n,r)= n!= (n-r)! P(8,4)= 8!= 8! =8(7)(6)(5)=1680 different passwords 8 - 4! 4! Exam Page 3 Find the answer to each of the following by first reducing the fractions as much as possible: a) P(490,2)= P(n,r)=n! (n-r)! P(490,2)= 490! = 490!= 490(498)=244, (490-2)! 488! b) C(670,665)= C(n,r)= n! r!(n-r)! C(670,665)= 670! = 670! = 670(669)(668)(667)(666)=1,108,399,190, 665!(670-665)! 665!5! 5(4)(3)(2)(1)
- 1.0 points
Instructor Comments
Find the answer to each of the following by first reducing the fractions as much as possible: a) P(490,2)= b) C(670,665)= In (a) you multiplied by 498 instead of 489.
Answer Key
Exam Page 4 Suppose A and B are two events with probabilities: P(Ac^ )=.20,P(B)=.30,P(A∩B)=.20. a) What is (A│B)? P(AlB)= P(AnB) = .20 =. P(B). b) What is (B│A)? P(Ac)= P(A)=1-P(Ac)=1-.20=. P(BlA)= P(AnB) = .20 =. P(A). Exam Page 5 In a manufacturing plant, three machines A, B, and C produce 45 %, 35 %, and 20 %, respectively, of the total parts production. The company's quality control department determined that 1.5 % of the parts produced by machine A, 2 % of the parts produced by machine B, and 1 % 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 A?
Suppose that 5 out of 11 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in which they are chosen is not important. Since we do not want to count all of the possible orderings, we use combinations. You didn't need the 6 on top. It cancelled as part of the 6! on the bottom.
Answer Key
Exam 4
Exam Page 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency? For six failures: n=8, x=6, p=. f(x)= n! px(1-p)(n-x)^ = x!(n-x)! f(x)= 8! .08^6 (1-.08)(8-6)^ =. 6!(8-6)! For seven failures: n=8, x=7, p=. f(x)= n! px(1-p)(n-x)^ = x!(n-x)! f(x)= 8! .08^7 (1-.08)(7-6)^ =. 7!(8-7)! For eight failures: n=8, x=8, p=. f(x)= n! px(1-p)(n-x)^ = x!(n-x)! f(x)= 8! .08^8 (1-.08)(8-8)^ =. 8!(8-8)!
A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency? The probability is: .000006209+.000000155+.000000002=.
Instructor Comments
The numbers are so small that with rounding your are fine.
Answer Key
Exam Page 2 Find each of the following probabilities: a. Find P(Z ≤ 1.27). P(Z<1.27)=. b. Find P(Z ≥ - .73). P(Z>-.73)=1-P(Z<-.73)=1-.23270=.
Policies Sold Per Day Probability, f(x)
Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data. E(x)=u=∑xf(x)= =0(.04)+1(.11)+2(.23)+3(.26)+4(.19)+5(.17)=2. The insurance salesperson can expect to sell 2.96 insurance policies per day. Variance: Var(x)=o^2 =∑(x-u)^2 f(x)= =(0-2.96)^2 (.04)+(1-2.96)^2 (.11)+(2-2.96)^2 (.23)+(3-2.96)^2 (.26)+(4-2.96)^2 (.19)+(5- 2.96)^2 (.17)=1. Standard Deviation: o=√o^2 = o=√o^2 =√1.8984=1. Exam Page 5 An archer is shooting arrows at a target. She hits the target 55% of the time. If she takes 14 shots at the target, what is the probability that she will hit the target exactly 7 times? n=14, x=7, p=. f(x)= 14! .55^7 (1-.55)(14-7)^ =. 14!(14-7)!
- 2.0 points
Instructor Comments
You can see in the Answer Key where you are off in the denominator, which throws your answer off.
An archer is shooting arrows at a target. She hits the target 55% of the time. If she takes 14 shots at the target, what is the probability that she will hit the target exactly 7 times?
Answer Key
Exam 5
Exam Page 1
Suppose that you take a sample of size 26 from a population that is not normally distributed. Can the sampling distribution of xx be approximated by a normal probability distribution? The sampling distribution of xx cannot be approximated by a normal probability distribution because if the population isn't normally distributed, the sample size has to be at least 30.
Exam Page 2
Suppose that you are attempting to estimate the annual income of 1200 families. In order to use the infinite standard deviation formula, what sample size, n, should you use? n < 0.05= N n < 0.05= 1200 n < 0.05(1200) n < 60 Sample size must be less than 60.
Exam Page 3
Exam 6
Exam Page 1
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug is given to 70 people who have fevers. It is found that the mean time that it takes for the fever to get back to normal for this test group is 400 minutes with a standard deviation of 85 minutes. Find the 95% confidence interval for the mean time that the drug will take to reduce all fevers for all people. Case 1: Large population and large sample size n=70 xx = 400 s=85 z=1. xx - z s <u< xx +z s √n √n 400 - 1.96 85 <u<400+1.96 85 √70 √ 380.0875<u<419.
Exam Page 2
A certain school has 380 male students. The school nurse would like to know how many calories the male students consume per day. So, she samples 65 male students and finds that the mean calorie consumption of the 65 is 1950 calories per day with a standard deviation of 200 calories per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the school. Case 3: Finite population N=300 n=65 xx = 1950 s=200 z=1. xx - z s √N-n<u< xx +z s √N-n √n √N- 1 √n √N- 1 1950 - 1.96 200 √ 300 - 65 <u<1950+1.96 200 √ 300 - 65 We want P(Z< - .07). From the standard normal table, we find: P(Z< - .07)=.47210. So there is a .47210 probability that the percentage of the sample that have more than two jobs is less than 7 %.
A certain school has 380 male students. The school nurse would like to know how many calories the male students consume per day. So, she samples 65 male students and finds that the mean calorie consumption of the 65 is 1950 calories per day with a standard deviation of 200 calories per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the school. The population is finite. So, we should use Case 3: Finite population. Use: In the statement of the problem, we are given: N=380 n=65 x ¯=1950 s= For a 95% confidence level, table 6.1 gives z=1. 1905.67 < μ< 1994.
1906.895<u<1993.
- 2.0 points
Instructor Comments
1950 - 1.96 200 √ 300 - 65 <u<1950+1.96 200 √ 300 - 65 √65 √ 300 - 1 √65 √ 300 - 1 300 should be 380.
Answer Key
Exam Page 3
