Stat Study Guide
Determine the area under the standard normal curve that lies to the right of (a) Z=−0.25, (b) Z=1.59, (c) Z=1.25, and (d) Z=−1.23.
RIGHT means values will be in the LOWER bound and E99 for UPPER bound, 0 then 1. USE NORMALCDF.
Determine whether the distribution is a discrete probability distribution.
Sum of probabilities must be equal to 1 and each probability must be between 0 and 1
What is the probability of obtaining ten tails in a row when flipping a coin? Interpret this probability.
The probability of obtaining ten tails in a row when flipping a coin is 0.00098 Consider the event of a coin being flipped ten times. If that event is repeated ten thousand different times, it is expected that the event would result in ten tails about 1010 time(s).
For the fiscal year 2007, a tax authority audited 1.84% of individual tax returns with income of $100,000 or more. Suppose this percentage stays the same for the current tax year. What is the probability that two randomly selected returns with income of $100,000 or more will be audited?
0.0184 x 0.0184 = 0.000339
Suppose that E and F are two events and that P(E)=0.9 and P(F|E)=0.2. What is P(E and F)?
0.18 (0.9 x 0.2)
Suppose that E and F are two events and that P(E and F)=0.1 and P(E)=0.2. What is P(F|E)?
0.50 (0.1/0.2)
Find the Z-scores that separate the middle 10% of the distribution from the area in the tails of the standard normal distribution.
1 - .10 = 0.9 0.9/2 = 0.45 use infvnorm formula and put both negative and positive numbers = -0.13 and 0.13
Find the Z-score such that the area under the standard normal curve to the right is 0.11.
1 - 0.11 = 0.89 then use infvnorm = 1.23
Find the probability P(Ec) if P(E)=0.17.
1 - 0.17 = 0.83
Find the value of zα. z0.18
1 - 0.18 = 0.82 then use infvnorm form = 0.92
The mean gas mileage for a hybrid car is 56 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.2 miles per gallon. (a) What proportion of hybrids gets over 61 miles per gallon? (b) What proportion of hybrids gets 50 miles per gallon or less? (c) What proportion of hybrids gets between 57 and 62 miles per gallon? (d) What is the probability that a randomly selected hybrid gets less than 46 miles per gallon?
use ncdf
Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8}. Suppose the outcomes are equally likely. Compute the probability of the event E="an odd number."
1, 3, 5 and 7 are odd numbers so 4/8 = 0.5
Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E="an even number less than 10."
2, 4, 6, 8 are even numbers less than 10 so 4/10 = 0.4
Let the sample space be S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose the outcomes are equally likely. Compute the probability of the event E={1, 4, 7, 9}.
4/10 (number of the amount of probabilities divided by the total number) = 0.4
According to a certain country's department of education, 44.4% of 3-year-olds are enrolled in day care. What is the probability that a randomly selected 3-year-old is enrolled in day care?
44.4/100 = 0.444 (Divide by 100)
An investment counselor calls with a hot stock tip. He believes that if the economy remains strong, the investment will result in a profit of $50,000. If the economy grows at a moderate pace, the investment will result in a profit of $10,000. However, if the economy goes into recession, the investment will result in a loss of $50,000. You contact an economist who believes there is a 30% probability the economy will remain strong, a 60% probability the economy will grow at a moderate pace, and a 10% probability the economy will slip into recession. What is the expected profit from this investment?
50,000 x .3 + 10,000 x .6 + (-50,000) x .1 = 16000
Determine the area under the standard normal curve that lies between (a) Z=−0.35 and Z=0.35, (b) Z=−2.47 and Z=0, and (c) Z=−1.45 and Z=−0.97.
LOWER BOUND THEN UPPER BOUND - use normcdf
A probability experiment is conducted in which the sample space of the experiment is S={12,13,14,15,16,17,18,19,20,21,22,23}. Let event E={13,14,15,16,17,18} and event F={17,18,19,20}. List the outcomes in E and F. Are E and F mutually exclusive?
List the outcomes in E and F. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. 17, 18 Are E and F mutually exclusive? D. No. E and F have outcomes in common.
A probability experiment is conducted in which the sample space of the experiment is S={6,7,8,9,10,11,12,13,14,15,16,17}. Let event E={7,8,9,10}. Assume each outcome is equally likely. List the outcomes in Ec. Find PEc.
List the outcomes in Ec. (numbers not in event E) A. Ec= 6, 11, 12, 13, 14, 15, 16, 17 P(E) = 0.333 (outcomes in event E divided by total outcomes) P(Ec) = 1 - 0.333 = 0.667
A probability experiment is conducted in which the sample space of the experiment is S={2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, event F={6, 7, 8, 9, 10, 11}, and event G={10, 11, 12, 13}. Assume that each outcome is equally likely. List the outcomes in F or G. Find P(F or G) by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule.
List the outcomes in F or G. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. F or G= 6, 7, 8, 9, 10, 11, 12, 13 B. Find P(F or G) by counting the number of outcomes in F or G. P(F or G)=0.667 (number of outcomes divided by total number) Determine P(F or G) using the general addition rule. Select the correct choice below and fill in any answer boxes within your choice. A. P(F or G) = 0.500 + 0.333 - 0.167 = 0.667 ( P(F or G) = P(F) + P(G) - P(F - G) )
A random sample of 80 high school seniors is obtained, and the individuals selected are asked to state their hair length.
No, this probability experiment does not represent a binomial experiment because the variable is continuous, and there are not two mutually exclusive outcomes.
Find the probability of the indicated event if P(E)=0.25 and P(F)=0.55. Find P(E or F) if P(E and F)=0.05.
P(E or F) = P(E) + P(F) - P(E n F) = 0.25 + 0.55 - 0.05 = 0.75
If P(E)=0.45, P(E or F)=0.50, and P(E and F)=0.10, find P(F).
P(E or F) = P(E) + P(F) - P(E n F) = 0.50 = 0.45 + 0.15 - 0.10
Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.25, and P(F)=0.51.
P(E or F) = P(E) + P(F) = 0.25 + 0.51 = 0.76
A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n=9, p=0.3, x≤3 (less than or equal to x - use cdf)
use BINOMCDF (2ND, VAR - scroll down)
A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n=10, p=0.65, x= 7 (single value - use pdf)
use BINOMPDF (2ND, VAR - scroll down)
A bag of 100 tulip bulbs purchased from a nursery contains 35 red tulip bulbs, 25 yellow tulip bulbs, and 40 purple tulip bulbs. (a) What is the probability that a randomly selected tulip bulb is red? (b) What is the probability that a randomly selected tulip bulb is purple? (c) Interpret these two probabilities.
(a) 35/100 = 0.35 (b) 40/100 = 0.4 (c) If 100 tulip bulbs were sampled with replacement, one would expect ABOUT 35 of the bulbs to be red and ABOUT 40 of the bulbs to be purple.
A standard deck of cards contains 52 cards. One card is selected from the deck. (a) Compute the probability of randomly selecting a club or diamond. (b) Compute the probability of randomly selecting a club or diamond or heart. (c) Compute the probability of randomly selecting a six or spade.
(a) = 0.500 (b) = 0.750 (c) = 0.308
Suppose that two cards are randomly selected from a standard 52-card deck. (a) What is the probability that the first card is a queen and the second card is a queen if the sampling is done without replacement? (b) What is the probability that
(a) If the sampling is done without replacement, the probability that the first card is a queen and the second card is a queen is 0.005 (b) If the sampling is done with replacement, the probability that the first card is a queen and the second card is a queen is 0.006
The data in the following table show the association between cigar smoking and death from cancer for 129,616 men. Note: current cigar smoker means cigar smoker at time of death. (a) If an individual is randomly selected from this study, what is the probability that he died from cancer? (b) If an individual is randomly selected from this study, what is the probability that he was a current cigar smoker? (c) If an individual is randomly selected from this study, what is the probability that he died from cancer and was a current cigar smoker? (d) If an individual is randomly selected from this study, what is the probability that he died from cancer or was a current cigar smoker?
(a) P(died from cancer)=0.006 (b) P(current cigar smoker)=0.051 (c) P(died from cancer and current cigar smoker)=0.001 (d) P(died from cancer or current cigar smoker)=0.056
Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable. (a) The number of points scored during a basketball game. (b) The amount of rain in City B during April.
(a) The random variable is discrete. The possible values are x=0, 1, 2,... (b) The random variable is continuous. The possible values are r≥0.
The data represent the number of driving fatalities for a certain area by age for male and female drivers.
(a) What is the probability that a randomly selected driver fatality who was female was 16 to 20 years old? The probability that a randomly selected driver fatality who was female was 16 to 20 years old is approximately 0.155 (b) What is the probability that a randomly selected driver fatality who was 16 to 20 was female? The probability that a randomly selected driver fatality who was 16 to 20 was female is approximately 0.298 (c) Is a victim of a fatal accident aged 16 to 20 more likely to be male or female? Choose the correct statement below. C. The driver is more likely to be male because the probability is greater than 0.5. Your answer is correct.
In a recent poll, a random sample of adults in some country (18 years and older) was asked, "When you see an ad emphasizing that a product is "Made in our country," are you more likely to buy it, less likely to buy it, or neither more nor less likely to buy it?" The results of the survey, by age group, are presented in the following contingency table. Complete parts (a) through (c).
(a) What is the probability that a randomly selected individual is at least 55 years of age, given the individual is less likely to buy a product emphasized as "Made in our country"? The probability is approximately 0.254 (b) What is the probability that a randomly selected individual is less likely to buy a product emphasized as "Made in our country," given the individual is at least 55 years of age? The probability is approximately 0.030 (c) Are 18- to 34-year-olds more likely to buy a product emphasized as "Made in our country" than individuals in general? No, less likely
According to a survey, the probability that a randomly selected worker primarily drives a car to work is 0.752. The probability that a randomly selected worker primarily takes public transportation to work is 0.097. Complete parts (a) through (d).
(a) What is the probability that a randomly selected worker primarily drives a car or takes public transportation to work? = 0.849 (0.752 + 0.097) (b) What is the probability that a randomly selected worker primarily neither drives a car nor takes public transportation to work? = 0.151 (1 - 0.849) (c) What is the probability that a randomly selected worker primarily does not drive a car to work? = 0.248 (1 - 0.752) (d) Can the probability that a randomly selected worker primarily walks to work equal 0.25? Why or why not? No. The probability a worker primarily drives, walks, or takes public transportation would be greater than 1.
Among 43- to 48-year-olds, 31% say they have written an editorial letter while under the influence of alcohol. Suppose three 43- to 48-year-olds are selected at random. Complete parts (a) through (d) below.
(a) What is the probability that all three have written an editorial letter while under the influence of alcohol? 0.0298 (.31^3) (b) What is the probability that at least one has not written an editorial letter while under the influence of alcohol? 0.9702 (1-.31^3) (c) What is the probability that none of the three have written an editorial letter while under the influence of alcohol? 0.3285 (.69^3) (d) What is the probability that at least one has written an editorial letter while under the influence of alcohol? 0.6715 (1-.69^3)
Suppose a life insurance company sells a $180,000 one-year term life insurance policy to a 25-year-old female for $340. The probability that the female survives the year is 0.999627. Compute and interpret the expected value of this policy to the insurance company.
-1 - 0.999627 = 0.000373 -$340 - $180,000 = -179,660 -$340 x 0.999627 = 339.87 - (-179,660) x 0.000373 = -67.01 - (-67.01) + 339.87 = 272.86
Impossible Event
-A probability that os equal to 0
Probability Model
-All probabilities must sum to 1 -All probabilities must be between 0 and 1
In a national survey college students were asked, "How often do you wear a seat belt when riding in a car driven by someone else?" The response frequencies appear in the table to the right. (a) Construct a probability model for seat-belt use by a passenger. (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else? Response Frequency Never - 113 Rarely - 332 Sometimes - 588 Most of the time - 1067 Always - 2761
Add numbers up and divide by total to get frequency table which is Response Probability Never - 0.023 Rarely - 0.068 Sometimes - 0.121 Most of the time - 0.220 Always - 0.568 Unusual event is less than 0.05. (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else? A. Yes, because P(never)<0.05.
Which of the following numbers could be the probability of an event?
Any number between 0 and 1 EXAMPLES: 0.01, 1.21, 0, 1, 0.26 NONEXAMPLES: -0.45
If a person spins a six-space spinner and then draws a playing card and checks its color, describe the sample space of possible outcomes using 1, 2, 3, 4, 5, 6 for the spinner outcomes and B, R for the card outcomes.
B1, B2, B3, B4, B5, B6, R1, R2, R3, R4, R5, R6
The following data represent the number of games played in each series of an annual tournament from 1931 to 2003. Complete parts (a) through (d) below.
Construct a discrete probability distribution by adding up all frequencies then divide the frequency by the total. Compute and interpret the mean of the random variable x. (STAT, EDIT then enter values in L1, which is x, and frequency values in L2, press STAT, CALC THEN HIT VAR-1 STATUS, list should be L1 and frequency should be L2, press CALCULATE) - Mean is Ex and standard deviation is ox.
E: A person living at least 70 years. F: The same person regularly handling venomous snakes.
E and F are dependent because regularly handling venomous snakes can affect the probability of a person living at least 70 years.
E: A randomly selected person coloring her hair black. F: A different randomly selected person coloring her hair blond.
E cannot affect F and vice versa because the people were randomly selected, so the events are independent.
Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. P(X>40)
If x is greater than the value then the value is the lower bound. use normalcdf = 0.9234
Determine the area under the standard normal curve that lies to the left of (a) Z=1.47, (b) Z=1.15, (c) Z=−0.27, and (d) Z=−0.17.
LEFT means values will be in the upper bound and -E99 for lower bound, 0 then 1. USE NORMALCDF.
E: The rapid spread of a cocoa plant disease. F: The price of chocolate.
The rapid spread of a cocoa plant disease could affect the price of chocolate, so E and F are dependent.
Find the Z-score such that the area under the standard normal curve to the left is 0.52.
USE infvnorm = 0.05
An experimental drug is administered to 190 randomly selected individuals, with the number of individuals responding favorably recorded.
Yes, because the experiment satisfies all the criteria for a binomial experiment.
Classical Method
a method of assigning probabilities that is appropriate when all the experimental outcomes are equally likely (The probability of having six girls in an six-child family is 0.015625.)
The word or in probability implies that we use the .......... Rule.
addition
Subjective probability
the individual's personal estimate of the chance of loss (According to a sports analyst, the probability that a football team will win the next game is 0.49.)
When two events are disjoint, they are also independent.
false
Two events E and F are ________ if the occurrence of event E in a probability experiment does not affect the probability of event F.
independent
Assume the random variable X is normally distributed, with mean μ=57 and standard deviation σ=5. Find the 15th percentile
invnormal (.15, 57, 5) = 51.82
Empirical Method
method for acquiring knowledge based on observation, including experimentation, rather than a method based only on forms of logical argument or previous authorities (On the basis of a survey of 1000 families with six children, the probability of a family having six girls is 0.0064.) (On the basis of clinical trials, the probability of efficacy of a new drug is 0.73.)
The word and in probability implies that we use the ________ rule.
multuplication
About 12% of the population of a large country is nervous around strangers. If two people are randomly selected, what is the probability both are nervous around strangers? What is the probability at least one is nervous around strangers? Assume the events are independent.
(a) The probability that both will be nervous around strangers is 0.0144 (0.12 x 0.12) (b) The probability that at least one person is nervous around strangers is 0.2256 (0.12 + 0.12 - 0.0144) (c) The probability that Emerson loses five ping-pong games in a row, but does not lose six in a row is 0.0061 (.40^5)(1-.40)