Probability
Of the twenty-two students in a classroom, ten are transfer students, seven are nursing students, four are AAS students and one student is undecided:
If one student is chosen randomly, find the probability that the student is not an AAS student.
P(not AAS) = 1 - P(AAS) = 1 - (4/22) = .818 or 81.8%
Your company plans to invest in a business venture. If you do so, there is a 35% chance you will lose $30,000, a 40% chance you will break even, and a 25% chance you will make $55,000. Find the expected value.
-$30000(.35) + $55000(.25) = $3,250
Using the #6 medical probability table.
Determine the sensitivity of the test.
Sensitivity = true + / has condition
52/(52+17) = 52/69 = .7536 or 75.4%
Using graph for #9, What is the average life expectancy for a person born in Britain in the year 2000?
78 years
Of the twenty-two students in a classroom, ten are transfer students, seven are nursing students, four are AAS students and one student is undecided:
If one student is chosen randomly, find the probability that the student is either transfer or nursing.
P(Transfer or Nursing) = P(T) + P(N) = (10/22) + (7/22) = 17/22 = .773 or 77.3%
The night watchman in a factory cannot guard both the safe in the back and the cash register in front. The safe contains $6000, while the register has only $1000. Tonight the guard fears a robbery; the probability that the thief will try the cash register is 0.8 and the probability the thief will try the safe is 0.2. If the guard is not present, the thief will take all the money. If the guard is present, the thief will go away empty-handed. Where should the guard be positioned in order to minimize the thief's gains?
$6000(.2) = $1200
$1000(.8) = $800
The guard should watch the back safe.
Using the #6 medical probability table.
Determine the specificity of the test.
specificity = true - / doesn't have
357/(12+357) = 357/369 = .9674 or 96.7%
Using graph for #9, Find the percent of change for the France graph from the years 1980 to 2016. Write a sentence describing what you found, in context.
1980 = 74.3 and 2016 = 82.3
percent change = difference/orginal
(82.3 - 74.3)/74.3 = .10767 or 10.8%
The average life expectancy in France increased 10.8% from 1980 to 2016.
Of the twenty-two students in a classroom, ten are transfer students, seven are nursing students, four are AAS students and one student is undecided:
If three students are chosen randomly, without replacement, find the probability that all three students are nursing students.
P(all 3 students Nursing) = (7/22) * (6/21) * (5/20) = .0227 or 2.3%
Annaleigh pays $750 for her yearly car insurance and actuaries determined that she has a 5% chance of being in a car wreck. Actuaries further determined that the insurance company can expect to pay out an average of $5000 per claim. What is the insurance companies expected profit from Annaleigh’s policy?
E = $5000(.05) = $250
The insurance's profit is $750 - $250 = $500 from her policy.
Using the #6 medical probability table.
Determine the PPV of the test.
PPV = true + / All positive
52/(52+12) = 52/64 = .8125 or 81.3%
Of the twenty-two students in a classroom, ten are transfer students, seven are nursing students, four are AAS students and one student is undecided:
If five students are chosen randomly, with replacement, find the probability that at least one student is a transfer student.
P(at least 1 transfer in 5 selections) = 1 - [P(not transfer)]^5 = 1 - (12/22)^5 = .9517 or 95.2%
Annaleigh pays $750 for her yearly car insurance and actuaries determined that she has a 5% chance of being in a car wreck. Actuaries further determined that the insurance company can expect to pay out an average of $5000 per claim. If the insurance company has 500 policies similar to that of Annaleigh’s, how much profit can the insurance company expect to make?
$500 * 500 policies = $250,000 profit for 500 policies
Using the #6 medical probability table.
Determine the NPV of the test.
NPV = true - / All negative
357/(17+357) = 357/374 = .9545 or 95.5%
Each question on a multiple-choice exam has four choices. One of the choices is the correct answer, worth 4 points, another choice is wrong but still carries a partial credit of 2 points, and the other two choices are worth 0 points.
If a student picks answers at random, what is the expected value of his or her score for a problem?
If the exam has 80 questions, what is his or her expected score?
E = 4(1/4) + 2(1/4) = 1.5 pts per question
1.5 * 80 = 120 points
80*4 = 320 points possible
So, 120/320 = 37.5% is the expected score
Using the #7 medical probability table.
What is the probability of a person testing negative given that the person does have the disease?
negative / has the disease
1000000 * .20 = 200000 This is the total number who has the disease
Given the sensitivity is 75%, true +/has the disease = .75
.75 = x/200000 --> 200000*(.75) = 150000 this is the number of true positives
200000 - 150000 = 50000 this is the number of negative results that have the disease
So, 50000/200000 = .25 or 25% false negatives