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1.A large industrial firm purchases several new word processors

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1.A large industrial firm purchases several new word processors at the end of each year. The exact number of word processors purchased depends on the frequency of the number of repairs in the previous year. Suppose that the number of word processors x that are purchased each year has the following probability distribution:
x 0 1 2 3
f(x) 1/10 3/10 2/5 1/5
The cost of the desired model is $1200 per processor and a discount of 50 X2 dollars is credited towards any purchase. There is also a fixed cost of $600 regardless of the number of processors purchased. How much can this firm expect to spend on new word processors?

2.The density function of a continuous random variable X, the total number of hours, in units of 100 hours, that a family runs a vacuum cleaner over a period of one year is described by the following probability density function:
a.Plot the PDF and CDF between the values of 0 < x < 2
b.What is the average (mean) number of hours per year that families run their vacuum cleaners
c.Find the 60th percentile number of hours per year that families run their vacuum cleaners
d.Find the probability that families run their vacuum cleaners between 50 and 120 hours per year

3.Find the probabilities for the following discrete random variables
a.Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability that 3 or less (P (x 3)) will go into overtime?
c.The probability that a person, living in a certain city, owns a dog is estimated to be 0.3. Find the probability that the tenth person randomly interviewed in that city is the sixth one to own a dog.
d.A local drugstore owner knows that, on average, 100 people per hour stop by his store. Find the probability that in a given 3 minute period, more than 5 (P(x > 5) people enter the store.
e.An electronic switching device occasionally malfunctions and may need to be replaced. It is known that the device is satisfactory if it makes, on average no more than .2 errors per hour. A particular five hour period is chosen as a “test” on the device. If no more than 1 error occurs, the device is considered satisfactory. What is the probability that a satisfactory device will be mis- diagnosed as “unsatisfactory” on the basis of the test?
f.A restaurant chef prepares a tossed salad containing, on average, 5 vegetables per day. Find the probability that the salad contains more than 5 vegetables on i) a given day and on ii)3 of the next 4 days?

4.An investment project is expected to earn $100,000 on average with a standard deviation of $50,000. Assume earnings from this investment project is a normal distribution
a.Determine the probability that the annual net cash flow will be negative
b.What is the annual net cash flow at a probability of 75%
c.Of the next 15 people that invest in this project, what is the probability that 5 or less people will result an annual net cash flow of at least $100,000
d.What would the standard deviation have to be if the mean is 100,000 and the probability that a random variable x is less than 109,000 is 60%; (P(x< 109,000) ) = .60?

6.A fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday, the demand for hamburgers is normally distributed with a mean 313 and a standard deviation 57; the demand for chicken sandwiches is normally distributed with a mean 93 and a standard deviation 22.
a.How many hamburgers must the restaurant stock to be 98% sure of not running out of stock on a given day?
b.Answer part a for chicken sandwiches.
c.If the restaurant stocks 400 hamburgers and 150 chicken sandwiches for a given day, what is the probability that it will run out of hamburgers or chicken sandwiches (or both) that day? Assume that the demand for hamburgers and the demand for chicken sandwiches are independent

Rajeevan Pillai :

The density function of the second question is missing. Send it.

Rajeevan Pillai :

Answer to Question 1

Probability distribution

x: 0 1 2 3

f(x): 1/10 3/10 2/5 1/5

The mean of X is E(X) = 0*(1/10)+1*(3/10)+2*(2/5)+3*(1/5) = 1.7

E(X^2) = 0^2*(1/10)+1^2*(3/10)+2^2*(2/5)+3^2*(1/5) = 3.7

Let Y be the cost purchasing X word processors.

Then Y = 600 + 1200X – 50X^2

E(Y) = E(600 + 1200X – 50X^2) = 600+1200E(X)-50E(X^2) = 600+1200*1.7-50*3.7 =2455

The firm is expected to spend $2,455 on word processors.

Rajeevan Pillai :

Answers to Question 4

μ=100,000

σ=50,000

(a)

P(X<0)=P(z<(0-100,000)/50,000) = P(z<-2)=0.0228

(b)

P(X<k)=0.75

P(z<(k-μ)/σ)=0.75

(k-μ)/σ = 0.6745

k=μ+0.6745σ=100,000+0.6745*50,000=133,724

(c)

P(X>=100,000)=P(z>=(100,000-100,000)/50,000)=P(z>=0)=0.5

Of the next 15 people that invest in this project, the probability that 5 or less people will result an annual net cash flow of at least $100,000 = C(15,0)*(0.5)^15 + C(15,1)*(0.5)^15 + C(15,2)*(0.5)^15 + C(15,3)*(0.5)^15 + C(15,4)*(0.5)^15 + C(15,5)*(0.5)^15 = 0.1509

(d)

μ=100,000

σ=?

P(X<109,000)=0.60

P(z<(109,000-100,000)/σ)=0.60

(109,000-100,000)/σ=0.2533

9,000/σ=0.2533

σ=9,000/0.2533 = 35,524

Rajeevan Pillai :

Answers to Question 6

μ1=313, σ1=57

μ2=93, σ2=22

(a)

P(X1<=k)=0.98

P(z<=(k- μ1)/ σ1)=0.98

(k- μ1)/ σ1=2.0537

μ1+2.0537 σ1=313+2.0537*57=430

(b)

P(X2<=k)=0.98

P(z<=(k- μ2)/ σ2)=0.98

(k- μ2)/ σ2=2.0537

μ2+2.0537 σ2=93+2.0537*22=138

(c)

P(X1>400)=P(z>(400-313)/57)=P(z>1.5263)=1-P(z<1.5263)=1-0.9365=0.0635

P(X2 > 150)= P(z>(150-93)/22)=P(z>2.5909)=1-P(z<2.5909)=1-0.9952=0.0048

P(run out of hamburgers or run out of chicken sand witches) = 0.0635+0.0048-0.0635*0.0048=0.0680

Rajeevan Pillai :

Questions 3(a) and 3(b) are mixed. Send them correctly.

Rajeevan Pillai :

3(c)

The probability = (C(9,5)*(0.3)^5*(0.7)^4) * 0.3=0.0221

Rajeevan Pillai :

3(d)

Rate per hour = 100

Rate per 3 minutes = (100/60)*3=5

The number of people X shopping in a 3 minute period can be assumed to be distributed as Poisson with mean 5.

P(X>5)=1-P(X=0)-P(x=1)- P(x=2)- P(x=3)- P(x=4)-P(x=5)=0.3840

Rajeevan Pillai :

3(e)

The number errors in 5 hour period is assumed to follow Poisson distribution with mean 1.

P(X>1)=1-P(X<=1) =1-P(X=0)-P(X=1)=1-0.3679-0.3679=0.2642

Rajeevan Pillai :

3(f)(i)

The number of vegetables X on a day follows Poisson distribution with mean 5.

P(X>5)=1=P(X<=5)=1-0.6160=0.3840

3f(ii)

The probability = C(4,3)*(0.3840^3)*(0.6160^1)=0.1395

Rajeevan Pillai :

Please Click ACCEPT IF I have helped you.

Rajeevan Pillai :

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JACUSTOMER-k280wrdb- :

The density function of a continuous random variable X, the total number of hours, in units of 100 hours, that a family runs a vacuum cleaner over a period of one year is described by the following probability density function:

f(x) =

x for 0< x < 1

2 – x for 1< x < 2

0 elsewhere

Plot the PDF and CDF between the values of 0 < x < 2

  1. What is the average (mean) number of hours per year that families run their vacuum cleaners
  2. Find the 60th percentile number of hours per year that families run their vacuum cleaners
  3. Find the probability that families run their vacuum cleaners between 50 and 120 hours per year
JACUSTOMER-k280wrdb- :

3.Find the probabilities for the following discrete random variables

a.Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability that 3 or less (P (x <= 3) are blemished?

JACUSTOMER-k280wrdb- :

b.The probability that a hockey game will go into overtime is .3; Of the next 12 hockey games, what is the probability that more than 3 (P(x >3)) will go into overtime?

Rajeevan Pillai :

Click the ACCEPT button if I have helped you. Positive feedback and bonus are appreciated.

JACUSTOMER-k280wrdb- :

σ=9,000/0.2533 = 35,524 or σ=9,000/0.2533 = 35,531

Rajeevan Pillai :

0.2533 is the value rounded to 4 decimal places. The value rounded to 9 decimal places is 0.253347103 . In the calculations, the value 0.253347103 is used.

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