MATH 110 Exam 1 (GRADED A) Questions and Answers | 100 OUT OF 100 | Latest Update, Exams of Nursing

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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.

Exam Page 2 Explain the difference between population and sample. "The entire number of items in a large group" would be defined as the population. (Module 1) The sample is then taken from the population by a researcher and is studied.The sample taken from the population is, in fact, the subset of the population. You need the population to get the sample and without the population, there can be no sample.

Instructor Comments

Very good definitions.

Answer Key

Exam Page 3 Look at the following data and see if you can identify any outliers: 65 71 55 69 3 77 67 70 246 61 277 3, 246, 277

Instructor Comments

Very good.

Answer Key

Explain the difference between population and sample. Population is the entire number of items in a large group. A sample is representative group from the population. Look at the following data and see if you can identify any outliers: 65 71 55 69 3 77 67 70 246 61 277

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

Chicken 10 Cheeseburger 6 Fish 18 Hamburger 8 Total 50 b) Make a relative frequency distribution for this data. Include relative percentages on this table. Types of Frequency Relative Relative Sandwiches Frequency Percentage Turkey 8 (8/50)= .16 (.16)100= 16% Chicken 10 (10/50)= .20 (.20)100= 20% Cheeseburger 12% Fish

Hamburger 8 (8/50)= .16 (.16)100= 16% Total 50 1 100% Exam Page 2 Consider the following data: 422 389 414 401 466 421 399 387 450 407 392 410 440 417 490 Find the 20th percentile of this data. 387,389,392,399,401,407,410,414,417,421,422,440,450,466, i=(p^ n^ = (20) 15= 3 100 ) 100 i= 392 is the 20th percentile of this data. Exam Page 3 Consider the following data: {29, 20, 24, 18, 32, 21} a) Find the sample mean of this data. x = ∑xi

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=

b) Find the mode of this set. {8,14,18,22,30,35,44,49} No mode (no number appears more than once).

Exam 3

Exam Page 1

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). P(AUB)=P(A)+P(B)-P(AnB) P(Bc)=> P(B)=1-P(Bc)=1-.45=. P(AUB)=.35+.55-.25=0. b) P(Ac^ ). P(Ac)=> P(A)=1-P(Ac)=> P(Ac)=1-P(A)= 1 - .35=. c) P(B). P(Bc)=> P(B)=1-P(Bc) P(B)=1-.45=.

Exam Page 2

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? 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)=

Machine A=P(A) Machine B=P(B) Machine C=P(C) Def=P(Def) P(A)=.45, P(B)=.35, P(C)=. P(DeflA)=.015, P(DeflB)=.02. P(DeflC)=. P(AlDef)= P(A)P(DeflA) = .45. =0. (P(A)P(DeflA))+(P(B)P(DeflB))+(P(C)P(DeflC)) .45.015+.35.02+.20. Exam Page 6 The probability that a certain type of battery in a smoke alarm will last 4 years or more is .65. The probability that a battery will last 7 years or more is .10. Suppose that the battery is 4 years old and is still working, what is the probability that the battery will last at least 7 years? P(E)=. P(F)=. (smoke alarm lasting more than 4 years) = E (smoke alarm lasting more than 7 years) = F If the smoke alarm has lasted 7 years, it would have lasted for 4 years already, so P(EnF)=. P(EnF)=. P(FlE)=P(EnF)=.10=. P(E). Exam Page 7 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. n=11 r= C(n,r)= n! r!(n-r)! C(11,5)= 11! = 11(10)(9)(8)(7)(6)= 5!6! 5(4)(3)(2)(1)

- 1.0 points

Instructor Comments

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=. 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?

c. Find P(-.09 ≤ Z ≤ .86). P(-.09<Z<.86)=P(Z<.86)-P(Z<-.09)=.80511-.46414=. Exam Page 3 A company manufactures a large number of rods. The lengths of the rods are normally distributed with a mean length of 3.7 inches and a standard deviation of .35 inches. If you choose a rod at random, what is the probability that the rod you chose will be: a) Less than 3.5 inches? u= 3.7, o=.35, x=3. z=x-u= 3.5-3.7=-. o. P(Z<-.57)=. b) Greater than 3.5 inches? u= 3.7, o=.35, x=3. z=x-u= 3.5-3.7=-. o. P(Z>-.57)=1-P(Z<-.57)=1-.28434=. c) Between 3.4 inches and 4 inches? u= 3.7, o=.35, x=3.4 x= z=x-u= 3.4-3.7=-.86 z=x-u= 4 - 3.7=. o .35 o. P(-.86<Z<.86)=P(Z<.86)-P(Z<-.86)=.80511-.19489=. Exam Page 4 A life insurance salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as follows:

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

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?

Suppose that in a large hospital system, that the average (mean) time that it takes for a nurse to take the temperature and blood pressure of a patient is 175 seconds with a standard deviation of 50 seconds. What is the probability that 35 nurses selected at random will have a mean time of 160 seconds or less to take the temperature and blood pressure of a patient? u=175, o=50, n=35, xx = oxx = o √n oxx = 50 = 8. √ z=xx - u= oxx z=xx - u= 160 - 175 = - 1. oxx 8. P(Z<-1.77)=. Exam Page 4 Suppose that in a very large city 7.2 % of the people have more than two jobs. Suppose that you take a random sample of 80 people in that city, what is the probability that 7 % or less of the 80 have more than two jobs? .

- 2.0 points

Instructor Comments

In your calculation of the standard deviation, you didn't get the square root. That throws the rest of your calculations off.

Answer Key

Suppose that in a very large city 7.2 % of the people have more than two jobs. Suppose that you take a random sample of 80 people in that city, what is the probability that 7 % or less of the 80 have more than two jobs? Now, we find the z-score:

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 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.

In a large city, the city supervisor wants to find the average number of aluminum cans that each family recycles per month. So, she surveys 21 families and finds that these 21 families recycle an average of 160 cans per month with a standard deviation of 45 cans per month. Find the 90 % confidence interval for the mean number of cans that all of the families in the city recycle per month. Case 2: Large population and small sample size n= 21 xx = 160 s=45 t= 1. 725 xx ±t s √n 160±1.725 45 √ 160±16. 143.0608<u<176. Exam Page 4 df=n-1=21-1= A doctor has a large number of patients and would like to know if his patients prefer to fill in forms electronically or prefer to hand write their forms. So, he surveys 95 patients and finds that 53 prefer electronic forms while 42 prefer hand written forms. Find the 95% confidence limit for the proportion of all patients that prefer the electronic forms. Case 1: P/p=p/n=53/95=.56 n=95 z=1. P± z √p(1-p) √n .56± 1.96 √.56(1-.56) √ .56±. .46 to. Exam Page 5 A shipment of 400 new blood pressure monitors have arrived. Tests are done on 55 of the new monitors and it is found that 10 of the 55 give incorrect blood pressure readings. Find the 90% confidence interval for the proportion of all the monitors that give incorrect readings. Case 2: P/p=p/n=10/55=.182 n=55 N= 400 z=1. P± z √p(1-p) √N-n √n √N- 1 .182± 1.645 √.182(1-.182) √400- 55 √55 √400- 1