### Discrete random variables

Discrete random variables

Question 1. (Discrete random variables)

The following table shows the frequency of the number of car accidents reported each day in a particular city over the last 50 days

Number of Accidents Frequency

1 23

2 15

3 7

4 3

5 2

a) Find the discrete probability distribution for the number of car accidents reported each day.

A: the number of accidents

P(A=1) = 23/50=0.46 per day

P(A=2)=15/50=0.3 per day

P(A=3)=7/50=0.14 per day

P(A=4)=3/50=0.06 per day

P(A=5)=2/50=0.04 per day

b) What is the mean (expected) number of accidents per day?

E(A)=(1*23+2*15+3*7+4*3+5*2)/50=1.92 per day

c) What is the standard deviation for the number of accidents per day?

Var(A)=(1-1.92)^2*(0.46)+(2-1.92)^2*(0.3)+(3-1.92)^2*(0.14)+(4-1.92)^2*(0.06)+(5-1.92)^2*(0.04)=1.1936

SD(A)=v1.1936=1.0925

d) What is the probability of fewer than three accidents of per day?

P(A<3)=P(A=1)+P(A=2)=0.46+0.3=0.76 per day

Question2.(Expected value and standard deviation of discrete random variables)Monthly sales have a mean of $25,000 and a standard deviation of $4,000. Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6,000.

a) Find the mean monthly profit.

?:Mean Monthly profit

?=$25000*30% -$6000=$1500

b) Find the standard deviation of monthly profits

Var(Profit)=Var(0.3*25000-6000)=0.3^2*25000=2250

SD(Profit)=v2250=47.4341649˜47.434

Question 3. (Binomial probability distributions)

Kobe Bryant, a professional basketball player in the NBA, has made 84% of his free throws during his career with the Los Angeles Lakers.

a) Calculate the probability that Bryant will make exactly three of his next five free throws.

P(successful)=0.84

P(failure)=1-0.84=0.16

P(three of the next five free throws)=5!/(3!*(5-3)!)*0.84^3*0.16^(5-3)=0.1517

b) Calculate the probability that Bryant will make less than five of his next six free throws.

P(less than five of his next six free throws)

=P(x<5)=1-P(x>=5)=1-(6!/5!(6-5)!)*0.84^5*0.16^(6-5))-(6!/(6!(6-6)!)*0.84^5-0.16^(6-5)=0.2472

c) Calculate the probability that Bryant will make four or five of his next six free throws.

P(four or five of his next six free throws)

=p(x=4)+P(x=5)=(6!/(4!(6-4)!)*0.84^4*0.16^(6-4))+(6!/(5!*(6-5)!)*0.84^5*0.16^(6-5))=0.5927

d) Do part (a) and (b) in MiniTab (or Excel) and show the result.

Question 4. (Poisson probability distributions)

According to a study, an average of six cell phone thefts is reported in Toronto per day. Assume that the number of reported cell phone thefts per day in the city follow the Poisson distribution.

a) What is the probability that exactly six cell phones will be reported stolen tomorrow?

) What is the probability that more than two cell phones will be reported stolen tomorrow?

c) What is the probability that four or five cell phones will be reported stolen tomorrow?

d) Perform part (a) and (b) in MiniTab (or Excel) and show the result.

Question 5. (Normal probability distributions)

According to College Board, the average room and board costs for a public four-year college in for the 2011-2012 academic year is $8,890. Assume that this cost follows

the normal distribution with a standard deviation of $975.

a) What is the probability that a randomly selected public four-year college has room and board costs greater than $9,000 during this year?

b) Perform part (a) in MiniTab (or Excel) and show the result.

Question 6. (Normal probability distributions)

According to Consumer Reports, the average annual out-of-pocket costs for an implant for Canadians who have dental insurance is $2,825 in 2011. Assume that this cost follows the normal distribution with a standard deviation of $270.

a) What is the probability that a randomly selected Canadian who has dental insurance will spend between $2,700 and $3,200 for an implant?

b) Perform part (a) in MiniTab (or Excel) and show the result.

Question 7. (Normal probability distribution)

A company’s manufacturing process uses 500 litres of water at a time. A “scrubbing” machine then removes most of a chemical pollutant before pumping the water into a nearby lake. To meet federal regulations the treated water must not contain more than 80 parts per million (ppm) of the chemical.Because there is a fine charged if regulations are not met, the company sets the machine to attain an average of 75 ppm

in the treated water. The machine’s output can be described by a Normal model with a standard deviation 4.2 ppm.

a) What percent of the batches of water discharged exceed the 80 ppm standard?

b) The company’s lawyers insist that not more than 2% of the treated water should be over the limit. In order to achieve this, to what mean should the company set the scrubbing machine? Assume the standard deviation does not change.Question 8. (Inverse normal probability calculations)In a weightlifting competition, the amount that the competitors can lift is normally distributed with µ = 196 kg and s = 11 kg. Only the top 20% of all competitors will be able to advance to the next phase of the competition.

a) What amount must a competitor lift in order to move into the next phase of the competition?

Question 8. (Inverse normal probability calculations)

In a weightlifting competition, the amount that the competitors can lift is normally distributed withµ = 196 kg and s = 11 kg. Only the top 20% of all competitors will be able to advance to the nextphase of the competition.

a) What amount must a competitor lift in order to move into the next phase of the competition?

b) Perform part (a) in MiniTab (or Excel) and show the result.

Question 9. (Normal approximation to the binomial distribution)

According to Nielsen, men account for 53% of tablet owners in 2011. A random sample of 75 tablet owners was selected.

a) Using the normal approximation to the binomial distribution, what is the probability that exactly 36 people from this sample were men?

b) Using the normal approximation to the binomial distribution, what is the probability that more than 34 people from this sample were men?

Question 10. (Normal approximation to the binomial distribution)

The owner of a pet store is trying to decide whether to discontinue selling specialty clothes for pets. She suspects that only 4% of the customers buy specialty clothes for their pets and thinks that she might be able to replace the clothes with more profitable items. Before making a final decision, she decides to keep track of the total number of customers for a day and whether they purchase specialty clothes.

a) The owner had 275 customers that day. Assuming this was a typical day for her store, what would be the mean and standard deviation of the number of customers who buy specialty clothes for their pet each day?

Binomial

n=275 so µ=np=275*0.04=11 customers

P=0.04 ?=vnpq=v275*0.04*0.96=3.25 customers

Q=0.96

b) Surprised by the high number of customers who purchased specialty pet clothing that day, the owner decided that her 4% estimate must have been too low. How many clothing sales would it have taken to convince you? Justify your answer.

Latest completed orders:

# | Title | Academic Level | Subject Area | # of Pages | Paper Urgency |
---|---|---|---|---|---|