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ECON IMPORTANT CONCEPTS EXAM 2

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A survey of a sample of business students resulted in the following information regarding the genders of the individuals and their selected major. Assign probabilities using the relative frequency method. Selected Major Gnd;mngt;mrkt;ot;tot Male;50 ;20 ;30;100 Female;30;20 ;50;100 Total; 80 ;40 ;80;200 a. What is the probability of selecting an individual who is majoring in Marketing?

0.2 with margin of 0 There are 40 individuals who are majoring in Marketing. The total number of individuals is 200. Hence, the probability of selecting an individual who is majoring in Marketing is 40/200 = 0.2.

The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample exactly two are male?

0.2090 The number of male students in the sample has a binomial probability distribution. The number of trials is n = 8. The probability of successes is p = 1 - 0.6 = 0.4. Therefore, the probability that the sample contains exactly two defective parts is given by (n!)/(x!(n−x)!) P^x (1−p)^(n-x) =(8!)/(2!(8−2)!) 0.42^2 (1−0.4)8−2 =0.2090.

combinations

The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N objects where order of selection is not important is called the rule for ORDER DOESN'T MATTER

The normal probability distribution is used with

a continuous random variable.

experiment

any process that generates well-defined outcomes

When the results of experimentation or historical data are used to assign probability values, the method used to assign probabilities is referred to as the _____ method.

relative frequency

Given that z is a standard normal random variable, what is the value of z if the area to the left of z is 0.2358?

-0.72 From the question, we know p(z≤z0)=0.2358. Checking the standard normal probability table, we know z0=−0.72.

Consider a uniform distribution ranging between 1 and 6. The probability density function for a value between 1 and 6 is

.2 f(x)= (1)/(b-a) = (1)/(6-1) = 0.2

For the sample of 20 Internet browser users, compute the expected number of Chrome users. (round your answers to the nearest thousandth)

4.074 with margin of .1 The expected number of Chrome users is equal to np = 0.2037×20 = 4.074.

addition law

P(A u B)= P(A)+P(B) - P(A intersect B)

The standard deviation of a normal distribution

cannot be negative

measures of association are:

covariance/ correlation coefficient

continuous random variable

may assume any numerical value in an interval or collection of intervals

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Compute P(D).

.23 with 0 margins The addition law implies that P(D) = P(A ∪ D) + P(A ∩ D) - P(A) = 0.4 = 0.6 + 0.03 - 0.4 = 0.23.

A production process produces 2.5% defective parts. A sample of five parts from the production process is selected. What is the probability that the sample contains exactly two defective parts?

0.0058 The number of defective parts in the sample has a binomial probability distribution. The number of trials is n = 5. The probability of successes is p = 0.025. Therefore, the probability that the sample contains exactly two defective parts is given by (n!)/(x!(n−x)!) P^x (1-p) ^(n-x) =(5!)/(2!(5−2)!) (0.025^2)(1−0.025)^(5−2)=0.0058.

x is a normally distributed random variable with a mean of 24 and a standard deviation of 6. The probability that x is less than 12 is

0.0228 p(x≤ 12) = p(z≤(12-24)/(6)=P(z≤-2) Checking the standard normal probability table, we find that p(z≤−2)=0.02275.

For the sample of 20 Internet browser users, compute the standard deviation for the number of Chrome users. (round your answers to the nearest thousandth)

1.8011 with margin of .1 The variance of the number of Chrome users is equal to np(1 - p) = 20×0.2037×(1 - 0.2037) = 3.2441262. The standard deviation is equal to sqrt of 3.2441262=1.801.

What is the maximum value of the middle 95% of cell phone prices? (round your answer to the nearest hundredth)

121.52 with margin of 1 The maximum value of the middle 95% has a percentile of 100 - (100 - 95)/2 = 97.5%. This implies that at this point, the cumulative probability of x is 0.975. We denote this minimum value as xH. Then we have P(x <= xH) = 0.975. Converting x to z, we have P(z <= (xH - 98)/12) = 0.975. As we learned in class, we need to find the z value in the cumulative probability table that corresponds to a cumulative probability that is the closest to 0.975. This z-value is 1.96. Hence, we have (xH - 98)/12 = 1.96. This implies that xH = 121.52.

If 7,218 of the Ahmadi cell phones were priced at least $119.00, __ thousand cell phones were produced by Ahmadi, Inc.. (round your answer to the nearest integer)

180 with margin of 5 We first compute the probability that x is at least 119, P(x >= 119). We know this is equal to 1 - P(x < 119). x follows a normal distribution so we first convert it to the standard normal random variable z. We have 1 - P(x < 119) = 1 - P(z <= (119 - 98)/12) = 1 - P(z <= 1.75). Checking the cumulative probability table for z, we find P(z <= 1.75) = 0.95994. Hence, we know P(x >= 119) = 1 - P(z <= 1.75) = 0.0401. We now know 0.0401 fraction of the total number of cell phones equals 7,218. This means the total number of cell phone equals 7218/0.0401 = 180,000. Therefore, the answer is 180 thousand cell phones.

Suppose you own one electronic store in Lincoln. You are interested in the number of laptops sold. The following table presents the probability distribution for number of laptops sold at the store, which is defined as x. xf(x) 2 0.2 4 0.4 8 0.4 What is the standard deviation of x?

2.4 with margin of .1 SD=var= square root of5.76 = 2.4

The probability that Pete will catch fish when he goes fishing is .88. Pete is going to fish for 3 days next week. Define the random variable x to be the number of days Pete catches fish. The expected number of days Pete will catch fish is

2.64 u=np=3*0.88=2.64

Assume that you have a binomial experiment with p = 0.4 and a sample size of 50. The mean and variance of this distribution (respectively) are

20 and 12 x has a binomial probability distribution with p = 0.4 and n = 50. Hence, E(x)= u = np = 50*0.4 = 20 and var(x) = np(1-p) = 50*.4*(1-.4) = 12

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Are A and B independent?

NO This is because P(A)P(B)=0.4×0.2=0.08, which is different from P(A∩B)=0.06.

Events A and B are mutually exclusive. Which of the following statements is also true?

P(A ∪ B) = P(A) + P(B)

if A and B are independent events with P(A) = .5 and P(A intersect B)=.12 then P(B)=

P(B)= .240 P(B)= (P(A intersect B))/ (P(a)) = .12/.5 = .24

permutations

The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N objects where order of selection is important is called the counting rule for ORDER MATTERS

A continuous random variable may assume

all values in an interval or collection of intervals.

event

collection of sample points (outcomes)

compliment of event A definition

defined to be the event consisting of all sample points that are not in A

The binomial probability distribution is used with a(n) _____ random variable. uniform discrete continuous intermittent

discrete

The probability that a continuous random variable takes any specific value

is equal to zero.

If P(A) = 0.7, P(B) = 0.6, P(A ∩ B) = 0, then events A and B are

mutually exclusive. A and B are mutually exclusive because P(A ∩ B) = 0. They are not independent because mutually exclusive events with non-zero probabilities cannot be independent. They are not complements of each other because P(A)+P(B)≠ 1.

For a continuous random variable x, the height of the function at x is

named the probability density function f(x).

standard normal distributions

normal distribution with a mean of 0 and a standard deviation of 1

The expected value for a binomial distribution is given by equation

the variance for the binomial distribution is given by equation

np(1-p)

random variable

numerical description of the outcome of an experiment

scale for probability values

probability values are always assigned on a scale from 0 to 1

process than generates well-defined experimental outcomes. on any single repetition or trial, one and only one of the possible experimental outcomes will occur and the outcome that occurs is determined completely by chance

random experiment

The area of the continuous uniform probability distribution is

rectangular

sample space

set of all experimental outcomes/ collection of all possible sample points

A method of assigning probabilities based upon judgment is referred to as the _____ method. relative subjective classical probability

subjective

events are mutually exclusive if:

the event has no sample points in common; the intersection must always be equal to 0

A negative value of z indicates that

the number of standard deviations of an observation is to the left of the mean.

conditional probability

the probability that one event happens given that another event is already known to have happened

uniformly distributed

whenever the probability is proportional to the interval's length

Larger values of the standard deviation result in a normal curve that is

wider and flatter.

Assume you have applied for two scholarships, a Merit scholarship (M) and an Athletic scholarship (A). The probability that you receive an Athletic scholarship is 0.18. The probability of receiving both scholarships is 0.11. The probability of getting at least one of the scholarships is 0.3. a. What is the probability that you will receive a Merit scholarship?

.23 From the question, we know P(A) = 0.18, P(A) = 0.18, P(A∩M)=0.11, and P(A∪M)=0.3. The addition law states that P(A∪M)=P(A)+P(M)−P(A∩M). This implies that P(M)=P(A∪M)+P(A∩M)−P(A)=0.3+0.11−0.18=0.23

.03168 Variance= np(1-p) =3*0.88 = 0.3168

The random variable x is known to be uniformly distributed between 50 and 90. The probability of x having a value between 70 to 80 is

.25 P(70 less than or equal to x less than or equal to 80) = (d-c)/(b-a)= (80-70)/(90-50)=0.25

A survey of a sample of business students resulted in the following information regarding the genders of the individuals and their selected major. Assign probabilities using the relative frequency method. Selected Major Gnd;mngt;mrkt;ot;tot Male;50 ;20 ;30;100 Female;30;20 ;50;100 Total; 80 ;40 ;80;200 b. What is the probability of selecting an individual who is majoring in Management, given that the person is female?

.3 with margin of 0 There are 30 individuals who are 1) majoring in Management and 2) female. There are 100 individuals who are female. Hence, the probability of selecting an individual who is majoring in Management, given that the person is female, is 30/100 = 0.3.

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Compute P(A | C).

.4 Using the definition of conditional probability, we know P(A | C) = p(a∩c)/p(c)=0.04/0.1=0.4.

A survey of a sample of business students resulted in the following information regarding the genders of the individuals and their selected major. Assign probabilities using the relative frequency method. Selected Major Gnd;mngt;mrkt;ot;tot Male;50 ;20 ;30;100 Female;30;20 ;50;100 Total; 80 ;40 ;80;200 d. What is the probability of selecting a male individual?

.5 There are 100 individuals who are male. The total number of individuals is 200. Hence, the probability of selecting a male individual is 100/200 = 0.5.

A survey of a sample of business students resulted in the following information regarding the genders of the individuals and their selected major. Assign probabilities using the relative frequency method. Selected Major Gnd;mngt;mrkt;ot;tot Male;50 ;20 ;30;100 Female;30;20 ;50;100 Total; 80 ;40 ;80;200 c. Given that a person is male, what is the probability that he is majoring in Management?

.5 There are 50 individuals who are 1) majoring in Management and 2) male. There are 100 individuals who are male. Hence, the probability of selecting an individual who is majoring in Management, given that the person is male, is 50/100 = 0.5.

If A and B are mutually exclusive events with P(A) = 0.25 and P(B) = 0.4, then P(A ∪ B) =

.65 Because A and B are mutually exclusive, we know P(A∩B)=0. Using the addition law, we know P(A u B)= P(a) + P(b) P(A∩B) =0.25+0.4−0=0.65.

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Compute the probability of the complement of C.

.9 P(C^c)= 1-P(c) = 1-.01 =0.9

range of probability values is

0 to 1

If A and B are mutually exclusive events with P(A) = 0.3 and P(B) = 0.5, then P(A ∩ B) =

0. When two events are mutually exclusive, the probability of their intersection is zero.

Web Browser market share analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of 2014, Google's Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website). For a randomly selected group of 20 Internet browser users, answer the following questions. Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser. (round your answer to the nearest four decimal places)

0.0243 with margin 0.01 The number of users who use Chrome among the 20 users follows a binomial probability distribution with p = 0.2037 and n = 20. Hence, the probability that x = 8 is equal to C(^20 over 8) 0.2037^8(1−0.2037)^20−8=0.0243.

0.038 The number of days in which Pete will catch fish has a binomial probability distribution. The number of trials is n = 3. The probability of success is p = 0.88. Therefore, the probability that Pete will catch fish in exactly one day is given by (n!)/(x!(n−x)!) P^x (1−p)^(n-x) =(3!)/(1!(3−1)!) 0.88^1 (1−0.88)3−1 =0.0380

If P(A) = 0.50, P(B) = 0.40 and P(A ∪ B) = 0.88, then P(B | A) =

0.04 The addition law states that P(A ∪ B) = P(A) + P(B) - P(A ∩ B). It implies that P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.5 + 0.4 - 0.88 = 0.02. Hence, we know P(B|A) = (P(A∩B)/P(A) =0.020.5=0.04.

The probability that Pete will catch fish when he goes fishing is .88. Pete is going to fish 3 days next week. Define the random variable x to be the number of days Pete catches fish. The probability that Pete will catch fish on one day or less is

0.040 The number of days Peter catches fish (x) has a binomial probability distribution. The number of trials is n = 3. The probability of successes is p = 0.88. We are trying to compute the probability that x <= 1, which is given by P(x less than or equal to 1)= P(x=0)+P(x=1)= (3!)/(0!(3-0)!) (0.88^0) (1-0.88^3-0)+ (3!)/(1!(3-1)!) (0.88^1) (1-0.88^3-1) = 0.001728 + 0.038016= 0.03974

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Compute P(A ∩ B).

0.06 with margin of 0 The multiplication law implies that P(A ∩ B) = P(A | B)P(B) = 0.3×0.2 = 0.06.

x is a normally distributed random variable with a mean of 6 and a variance of 4. The probability that x is greater than 8.75 is

0.0838

If A and B are independent events with P(A) = 0.4 and P(B) = 0.25, then P(A ∩ B) =

0.1 Because A and B are independent, we know P(A∩B)=P(A)P(B)=0.4×0.25=0.1.

If P(A) = 0.62, P(B) = 0.47, and P(A ∪ B) = 0.88, then P(A ∩ B) =

0.21 The addition law states that P(A ∪ B) = P(A) + P(B) - P(A ∩ B). It implies that P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.62 + 0.47 - 0.88 = 0.21.

x is a normally distributed random variable with a mean of 7 and a standard deviation of 2. The probability that x is between 6.48 and 7.56 is

0.2128 p(6.48≤ x ≤ 7.56) =p((6.48-7)/2) =p(z ≤ 0.28) - p(z ≤ -0.26) . Checking the standard normal probability table, we find that p(z≤0.28)=0.61026 and p(z ≤−0.26)=0.39743. Hence, p(6.48≤x≤7.56)=0.61026−0.39743=0.21283.

Compute the probability that no more than 3 of the 20 Internet browser users use Chrome as their Internet browser. (round your answer to the nearest four decimal places)

0.3955 with margin 0.01 The probability that x = 0, 1, 2, or 3 is equal to C(^20 over 0) (.2037^0) ((1-0.2037)^20-0) +C(^20 over 1) (.2037^1) ((1-0.2037)^20-1) +C(^20 over 2) (.2037^2)((1-0.2037)^20-2) +C(^20 over 3) (.2037^3) ((1-0.2037)^20-3) =.0105+0.0538+0.1307+0.2005=0.3955.

What is the probability that x is between 3 and 6, given the probability density function below? f(x)=(1/(2 square root of 2*pie)) e ^(-.5((x-3)/2)^2

0.4332 Comparing the probability density function in the question with the probability density function for a normal distribution, we find that x follows a normal distribution with u=3 and SD=2. This question asks you to compute P(3 <= x <= 6). We know this is equal to P(x <= 6) - P(x <= 3). We first convert x to the standard normal random variable z. We have P(x <= 6) - P(x <= 3) = P(z <= (6 - 3)/2) - P(z <= (3 - 3)/2) = P(z <= 1.5) - P(z <= 0). Checking the cumulative probability table for z, we find P(z <= 1.5) = 0.93319 and P(z <= 0) = 0.5. Hence, we know P(x <= 6) - P(x <= 3) = P(z <= 1.5) - P(z <= 0) = 0.93319 - 0.5 = 0.43319.

Assume you have applied for two scholarships, a Merit scholarship (M) and an Athletic scholarship (A). The probability that you receive an Athletic scholarship is 0.18. The probability of receiving both scholarships is 0.11. The probability of getting at least one of the scholarships is 0.3. d. What is the probability of receiving the Athletic scholarship given that you have been awarded the Merit scholarship? (round your answer to the nearest four decimal places)

0.4783 with margin 0.01 P(A|M) = (P(A∩M)/(P(M)) =0.11/0.23=0.4783.

What is the probability of receiving the Merit scholarship given that you have been awarded the Athletic scholarship? (round your answer to the nearest four decimal places)

0.6111 margin with 0.01 P(M|A) = (P(A∩M))/(P(A))=0.110.18=0.6111.

Which of the following are continuous random variables? The weight of an elephant The time to answer a questionnaire The number of floors in a skyscraper The square feet of countertop in a kitchen

1, 2, and 4 only

4 properties of a binomial experiment

1. The experiment consists of a sequence of n identical trials 2. Two outcomes, success and failure, are possible on each trial 3. The probability of a success, denoted by p, does not change from trial to trial (stationary assumption) 4. The trials are independent

two properties of a poisson experiment

1. The probability of an occurrence is the same for any two intervals of equal length. 2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

requirements for assigning probabilities (2)

1. probability assigned to each experimental outcome must be between 0 and 1, inclusively 2. the sum of the probabilities for all experimental outcomes must equal 1

5.2 with margin 0.1 u=E(x)=sum(x(f(x)))= 2*.2+4*.4+8*.4=.4+1.6+3.2=5.2

5.76 with margin .1 var(x)= sum(x-u)^2(f(x))= (2-5.2)^2 *.2 + (4-5.2)^2 *.4 +(8-5.2)^2 *.4 = 2.048 +0.5760+3.136=5.76

The ages of students at a university are normally distributed with a mean of 21. What percentage of the student body is at least 21 years old?

50% For any normal random variable x, the probability that x is larger (smaller) than its mean is 50%.

Which of the following is not a characteristic of the normal probability distribution? 99.72% of the time the random variable assumes a value within plus or minus 1 standard deviation of its mean. The mean is equal to the median, which is also equal to the mode. Symmetry The total area under the curve is always equal to 1.

99.72% of the time the random variable assumes a value within plus or minus 1 standard deviation of its mean.

When the assumption of equally likely outcomes is used to assign probability values, the method used to assign probabilities is referred to as the _____ method.

Classical

Which of the following statements is always true? I. −0.5≤P(Ei)≤0.5. II. P(A)=1−P(A^C). III. P(A)+P(B)=1. IV. ∑Pi>1.

II.

NO A and M are not independent. This is because P(A)P(M)=0.18×0.23=0.0414≠0.11=P(A∩M).

NO A and M are not mutually exclusive because P(A∩M)=0.11≠0.

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Are A and B mutually exclusive?

NO This is because P(A ∩ B) is not zero.

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Are A and C mutually exclusive?

NO This is because P(A ∩ C) is not zero.

Which of the following is not a property of a binomial experiment? Each outcome can be referred to as a success or a failure. The probabilities of the two outcomes can change from one trial to the next. The experiment consists of a sequence of n identical trials. The trials are independent.

The probabilities of the two outcomes can change from one trial to the next.

You are given the following information on Events A, B, C, and D. P(A) = 0.4 P(A ∪ D) = 0.6 P(B) = 0.2 P(A | B) = 0.3 P(C) = 0.1 P(A ∩ C) = 0.04 P(A ∩ D) = 0.03 Are A and C independent?

YES This is because P(A)P(C)=0.4×0.1=0.04, which is the same as P(A∩C)=0.04.

The uniform probability distribution is used with

a continuous random variable.

In statistical experiments, each time the experiment is repeated

a different outcome might occur

probability of a random variable

assuming a value within some given interval is defined by 1. area under the graph 2. probability density function of interval numbers

A probability distribution showing the probability of x successes in n trials, where the probability of success does not change from trial to trial, is termed a

binomial probability distribution.

Which of the following is a characteristic of the standard normal probability distribution? has a standard deviation of 0. can have mean of any numerical value. needs to have a mean of 0. must have a standard deviation of 1 and a mean of 0.

can have mean of any numerical value

Two events with nonzero probabilities

can not be both mutually exclusive and independent

union of events A and B is defined by

containing all sample points that are in A or B or both, the union of both events is denoted by A u B

A random variable that may take on any value in an interval or collection of intervals is known as a _____ random variable. continuous multivariate discrete interval

continuous

An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable in this experiment is a _____ random variable. continuous discrete mixed type multivariate

continuous

what is this an example of?? the weight of an object is an example of a...

continuous random variable

Cell phones with prices of at least $81.80 will get a free gift. __% of of the cell phones will be eligible for the free gift. (round your answer to the nearest hundredth)

90.32 or 91.15 with margin of 1 This question asks you to compute P(x >= 81.8). We know this is equal to 1 - P(x < 81.8). x follows a normal distribution so we first convert it to the standard normal random variable z. We have 1 - P(x < 81.8) = 1 - P(z <= (81.8 - 98)/12) = 1 - P(z <= -1.35). Checking the cumulative probability table for z, we find P(z <= -1.35) = 0.08851. Hence, we know P(x >= 81.8) = 1 - P(z <= -1.35) = 0.9115. This is equal to 91.15%.

An experiment consists of four outcomes with P(E subscript 1) = 0.2, P(E subscript 2) = 0.25, and P(E subscript 3) = 0.05. The probability of outcome E subscript 4 is

0.5 The sum of the probabilities of all experimental outcomes is 1. Hence, P(E subscript 4) = 1 - P(E subscript 1) - P(E subscript 2) - P(E subscript 3) = 1 - 0.2 - 0.25 - 0.05 = 0.5.

For a standard normal distribution, the probability of z≤0 is

0.5.

An experiment consists of three steps. There are five possible results on the first step, two possible results on the second step, and three possible results on the third step. The total number of experimental outcomes is

30 The total number of experimental outcomes equals the product of the numbers of experimental outcomes in the three steps: 5×2×3=30.

A college plans to interview 7 students for possible offer of graduate assistantships. The college has three assistantships available. How many groups of three can the college select?

35 In this question, the order of selection does not matter because all three assistantships are identical. Hence, the number of groups of three that can be selected equals C N/n =C 7/3= (7!)/(3!(7-3)!= 35

The probability distribution for the daily sales at Michael's Co. is given below. Daily Sales(In $1,000s);Prob 40 ;0.1 50 ;0.4 60 ;0.3 70 ;0.2 The expected value of daily sales is __.

56 u= 40* 0.1 +50*0.4 + 60*0.3 + 70*0.2= 56

What is the minimum value of the middle 95% of cell phone prices? (round your answer to the nearest hundredth)

74.48 with margin of 1 The minimum value of the middle 95% has a percentile of (100 - 95)/2 = 2.5%. This implies that at this point, the cumulative probability of x is 0.025. We denote this minimum value as xL. Then we have P(x <= xL) = 0.025. Converting x to z, we have P(z <= (xL - 98)/12) = 0.025. As we learned in class, we need to find the z value in the cumulative probability table that corresponds to a cumulative probability that is the closest to 0.025. This z-value is -1.96. Hence, we have (xL - 98)/12 = -1.96. This implies that xL = 74.48.

Twenty percent of the students in a class of 400 are planning to go to graduate school. The standard deviation of this binomial distribution is

8 We know this is a binomial distribution with n = 400 and p = 0.2. Then, the variance is give by np(1 - p) = 400×0.2×(1−0.2)=64. Hence, the standard deviation is sqrt 64=8.

The probability distribution for the daily sales at Michael's Co. is given below. Daily Sales(In $1,000s);Prob 40 ;0.1 50 ;0.4 60 ;0.3 70 ;0.2 the variance of daily sales is

84 Variance= (40-56)^2 *0.1 + (50-56)^2 *0.4 + (60-56)^2 * 0.3 + (70-56)^2 * 0.2 = 84

The random variable x is known to be uniformly distributed between 70 and 100. The expected value (mean) and variance of x (respectively) are

85 and 75 For a uniform probability distribution with a = 70 and b = 100. The expected value is (a + b)/2 = (70 + 100)/2 = 85. The variance is ((b-a)^2)/12=((100−70)^2)/12=(30^2)/12=75.

The continuous uniform and normal distributions

are all continuous probability distributions.

For the standard normal probability distribution, the area to the right of the mean is

0.5

If a coin is tossed three times, the likelihood of obtaining three heads in a row is __. (Assign probabilities using the classical method.)

.125 This is a 3-step experiment with each step having two possible results. Therefore, there are, in total, 2×2×2=8 experimental outcomes. It means that the probability of seeing any particular outcome, such as three heads in a row, equals 1/8 = 0.125, if we assign probabilities using the classical method.

if P(A) = .4 and P(B I A)= .35 and P(AuB)=.69 then P(B)= ????

.43 we know P(A intersect B) = P(B I A)P(A)=.35*.4 = .14 if addition law is P(A u B)= P(A)+P(B) - P(A intersect B) then it implies that P(B)=P(A u B) +P(A intersect B) - P(A) = .69+.14-.4=.43

The random variable x is known to be uniformly distributed between 70 and 100. The probability of x having a value between 80 to 95 is

.5 x has a continuous uniform probability distribution between a = 70 and b = 100. The probability that it is between c = 80 and d = 95 is given by P(c less than or equal to x less than or equal to d)= (min(d,b)-max(a,c))/(b-a)= (min(95,100)-max(70,80))/(100-70) =15/30 =0.5.

Events A and B are mutually exclusive with P(C) = 0.35 and P(B) = 0.25. Then, P(B superscript c) =

.75 P(B superscript c) = 1 - P(B) = 1 - 0.25 = 0.75.

If A and B are mutually exclusive events with P(A) = 0.295, P(B) = 0.32, then P(A | B) = 0.295

0 If A and B are mutually exclusive, then P(A∩B)=0. Hence, P(A I B)=(P(A∩B))/P(B)=0.

Three percent of the customers of a mortgage company default on their payments. A sample of five customers is selected. What is the probability that exactly two customers in the sample will default on their payments?

0.0082 The number of customers who will default on their payments in the sample has a binomial probability distribution. The number of trials is n = 5. The probability of successes is p = 0.03. Therefore, the probability that the number of customers who will default on their payments equals 2 (x = 2) is given by (n!/(x!(n-x)!) (p^x) (1-p)^n-x = (5!)/(2!(5-2)!) (0.03^2) ((1-0.03)(^5-2)) = 0.0082

The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages (x) in the city has the following probability distribution. x f(x) 0 0.80 1 0.15 2 0.04 3 0.01 The mean and the standard deviation for the number of electrical outages (respectively) are

0.26 and 0.577 u=∑xf(x)=0*.8+1*.15+2*.04+3*.01= .26 Variance=∑(x-u)^2(f(x))= (0−0.26)2×0.8+(1−0.26)2×0.15+(2−0.26)2×0.04+(3−0.26)2×0.01=0.3324. SD= square root of Variance

If A and B are independent events with P(A) = 0.38 and P(B) = 0.55, then P(A | B) =

0.38 If A and B are independent, then P(A) = P(A|B) = 0.38

The probability that it rains on any given day is 0.1. Assume that whether it rains on one day is independent of whether it rains on another day. What is the probability that it rains on at least one day in a week?

0.5217 The number of rainy days in a week has a binomial probability distribution. The number of trials is n = 7. The probability of successes is p = 0.1. We are trying to find the probability that the number of rainy days is greater than or equal to 1, f(1) + f(2) + ... + f(7). We note that this probability equals 1 - f(0), where f(0) is the probability that there are no rainy days in the week. Hence, we just need to compute f(0) first: F(0) = (n!)/(x!(n-x)!) (p^x) (1-p^n-x) = (7!)/(0!(7-0)!) (0.1^0) (1-0.1^7-0) = 0.4783 Then, we have f(1) + f(2) + ... + f(7) = 1 - f(0) = 0.5217.

If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A ∪ B) =

0.68 Because A and B are independent, we know P(A∩B)=P(A)P(B)=0.2×0.6=0.12. Then, the addition law implies that P(A∪B)=P(A)+P(B)−P(A∩B)=0.2+0.6−0.12=0.68.

If P(A) = 0.58, P(B) = 0.44, and P(A ∩ B) = 0.25, then P(A ∪ B) =

0.77 The addition law implies that P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.58 + 0.44 - 0.25 = 0.77.

z is a standard normal random variable. The p(−1.96≤z≤1.4) equals

0.8942 P(-1.96≤ z ≤ 1.4)= p(z≤1.4) - p(z ≤ -1.96 Checking the standard normal probability table, we find that p(z≤1.4)=0.91924 and p(z≤−1.96)=0.02500. Hence, p(−1.96≤z≤1.4)=0.91924−0.02500=0.89424.

Given that z is a standard normal random variable, what is the probability that z≤1.46?

0.92785 Checking the standard normal probability table, we find that p(z≤1.46)=0.92785.

Given that z is a standard normal random variable, what is the probability that z≥−2.12?

0.9830 p(z greater than or equal to -2.12)=1-p(z greater than or equal to -2.12) Checking the standard normal probability table, we find that p(z≤−2.12)=0.01700. Hence, p(z≥−2.12)=1−0.01700=0.98300.

The sum of the probabilities of two complementary events is

probabilities of a normal probability distribution

1. highest point on the normal curve is at its mean, which is also the median and mode 2. the mean can be any numerical value: negative, zero, or positive 3. the standard deviation determines the width of the curve- larger values result in wider flatter curves 4. probabilities for the normal random variable are given by areas under the curve. the total area under the curve is 1. to the right (left of the mean) is .5

A continuous random variable is uniformly distributed between a and b. The probability density function between a and b is

1/(b-a

Random variable x has the probability function f(x) = X/6, for x = 1, 2 or 3. The expected value of x is

2.333 u= 1/6*1 + 2/6*2 + 3/6*3=2.333

The average price of cell phones manufactured by Ahmadi, Inc. is $98 with a standard deviation of $12. Furthermore, it is known that the prices of the cell phones manufactured by Ahmadi are normally distributed. __% of cell phones produced by Ahmadi, Inc. will have prices of at least $120.20. (round your answer to the nearest hundredth)

3.22 with margin of 1 This question asks you to compute P(x >= 120.2). We know this is equal to 1 - P(x < 120.2). x follows a normal distribution so we first convert it to the standard normal random variable z. We have 1 - P(x < 120.2) = 1 - P(z <= (120.2 - 98)/12) = 1 - P(z <= 1.85). Checking the cumulative probability table for z, we find P(z <= 1.85) = 0.96784. Hence, we know P(x >= 120.2) = 1 - P(z <= 1.85) = 0.0322. This is equal to 3.22%.

specified interval of time or space

A Poisson distributed random variable is often useful in estimating the number of occurrences

bivariate probability distribution

A probability distribution involving two random variables.

Joe's Record World has two stores and sales at each store follow a normal distribution. For store 1, μ = $2,000 and σ = $200 per day; for store 2, μ = $1,900 and σ = $300 per day. Which store is more likely to have a day's sales in excess of $2200?

Store 1 and Store 2 are equally likely For store 1, the probability that a day's sales are in excess of $2200 is given by p(x≥2200)=1−p(z≤2000−2200/200)=1−p(z≤−1). For store 2, the probability that a day's sales are in excess of $2200 is given by p(x≥2200)=1−p(z≤1900−2200/300)=1−p(z≤−1). Hence, store 1 and store 2 are equally likely to have a day's sales in excess of $2200.

what is this an example of? the number of customers that enter a store during one day

a discrete random variable

probability distribution defined by what function

denoted by probability function f(x)- provides the probability for each value of the random variable

probability distribution

describes how probabilities are distributed over the values of the random variable and their associated probabilities

independent variables

do not change because another variable has changed, completely independent of one another.

sample point

each individual outcome of an experiment is

what is the probability of an event equal to?

equal to the sum of the probabilities of the sample points in the event

Which of the following is a characteristic of an experiment where the binomial probability distribution is applicable? The probabilities of the outcomes changes from one trial The trials are dependent on each other Each trial has more than two possible outcomes Exactly two outcomes are possible on each trial

exactly two outcomes are possible on each trial

For any continuous random variable, the probability that the random variable takes a value less than zero

is any number between zero and one.

For a uniform probability density function, the height of the function

is the same for each value of x.

entire family of normal probability distributions defined by:

mean, standard deviation

expected value

measure of the average value of a random variable (MEAN) a weighted average of the values of a random variable, where the probability function provides weights

discrete random variables

my assume either a finite number of values or an infinite sequence of values

numerical measure of the likelihood an event occurs

probability

The function that defines the probability distribution of a continuous random variable is a

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