Business Mathematics_2

Normal Distribution Problems

Example 1

• The average daily sales of 500 branch offices was 150 thousand and the S.D Rs.15 thousand . Assuming the distribution to b normal, indicate how many branches have sales between

Rs. 120 thousand and Rs. 145 thousand

Rs. 140 thousand and Rs. 165 thousand

1) P(120<=X<=145)=p(-2<=Z<=-.33)

=0.4772-0.1293

=0.3479

Expected number of branches

=0.3479 X 500 = 174

2) 0.5899 x 500=295

Example 2

• There are 600 business students in the graduate department of a university, and the probability for any student to need a copy of a particular textbook from the university library on any day is 0.05 . How many copies of the book should be kept in the university library so that the probability may greater than 0.90 that none of the students needing a copy from the library has to come Back disappointed ?

( use normal approximation to the binomial )

• Mean= np= 600 x 0.05=30

• Variance=npq= 600 x 0.05 x 0.95

• = 28.5

Example 3

• As a result of tests on 20,000 electric fans manufactured by a company, it was found that lifetime of the fans was normally distributed with an average life of 2040 hours and standard deviation of 60 hours. On the basis of the information, estimate the number of fans that is expected to run for (a) more than 2150 hours and (b) less than 1960 hours.

A) X =2150, mean=2040 S.D=60

Z=1.833

Area to the right of 1.833=0.5-0.4664

=0.0366

The no. of fans=0.0366 x 20000=672

B) The no. of fans =1836

Example 4

• A complex television component has 1000 joints by machine which is known to produce on average, one defect in forty. The components are examined, and faulty soldering corrected by hand. If components requiring more than 35 corrections are discarded, what proportion of the components will be thrown away?

Normal as limiting case of Poisson distribution

The average no. of defects=1000 x 0.4=25

S.D=sqrt(25)=5

Variance of Poisson distribution = root of mean

Z=(35-25)/5 = 2

Area=0.5-0.4772=0.0228 2.28%

Example 5

• The manager of a small postal substation is trying to quantify the variation in the weekly demand for mailing tubes. The information from quality control department is that this demand is normally distributed. Further from the earlier records she knows that on average 100 tubes are purchased weekly and that 90% of the time, weekly demand is below 115.

• The manager wants to stock enough mailing tubes each week so that the chance of running out of tubes is not higher than 0.05. Assuming you in place of manager decide the stock level.

• μ= 100 σ=?

• P(x<115)=> σ)=0.90

• =p(z<1.28)

• 115-100)/ σ = 1.28

• σ = 11.72

• P(x>?)=0.05 = P(z > (?-100)/11.72)=0.05=

• P(z>1.64)

• (?-100)/11.72=1.64

• ?=119.22

• The lowest suitable stock level is 120 tubes.

Example 6

• The Gilbert Machinery Company has received a big order to produce electric motors for a manufacturing company. In order to fit in its bearing, the drive shaft of the motor must have a diameter of 5.1 + 0.05 (inches). The company’s purchasing agent realizes that there is a large stock of steel rods in inventory with a mean diameter of 5.07” and a S.D of 0.07’. Assume that you are the purchase manager of the company. Are you going to prepare the purchase order?

μ= 5.07 σ=0.07

P(5.05< x <5.15)=

P(-.029 < z < 1.14)=

0.1141+.3729= 0.4870

Example 7

• The managing director of spiffy Lube auto lubrication Ltd. has asked his purchase manager to revise his policy on ordering grease gun cartridges. Currently, he orders 110 cartridges per week, but he runs out of cartridges 1 out of every 4 weeks. He know that on average, the shop uses 95 cartridges per week. He is also willing to assume that demand for cartridges is normally distributed.

• If the manager wants to order enough cartridges so that his chance of running out during any week is no greater than 0.2, revise purchasing policy.

μ= 95 σ=?

P(x>110)=P(z>((110-95)/ σ))=0.25 =P(z>0.67)

(110-95)/ σ)=0.67

σ=22.39

P(x>?)=0.2 = P(z>0.84)

(?-95)/22.39 =0.84

?=13.81

Example 8

• Harley Davidson, Director of quality control for the Kyoto Motor company, is conducting his monthly spot check of automatic transmissions. In this procedure 10 transmissions are removed from the pool of components and are checked for manufacturing defects. Historically, only 2% of the transmissions have such flaws. (assume that flaws occur independently in different transmissions)

• If the sample contains more than 2 transmissions with manufacturing flaws, the company has to revise its contract with King Motors Ltd. with 10 % discount. What is the chance of revising the contract?

• What is the chance of not revising the contract?

• P(more than 2 flaws)= 1 –

[ p(0 flaw)- P(1 flaw)-P(2 flaws)]

=0.0009 (revising )

p=0.02 n=10

q=1-p = 0.98

Binomial distribution = ⁿc_x^x q p^n-x

P(0 flaw)= ¹⁰c₀ p⁰ q^10-0 =0.8171 (not revising)

P(1 flaw)= ¹⁰c₁ p¹ q^10-1 =0.1667

P(2 flaws)= ¹⁰c₂ p² q^10-2 =0.0153

Example 9

• Diane Bruns is the mayor of a large city. Lately, she has become concerned about the possibility that large numbers of people who are drawing unemployment checks are secretly employed. Her assistants estimate that 40% of unemployment beneficiaries fall into this category, but Ms. Bruns is not convinced. She asks one of her aids to conduct a quiet investigation of 10 randomly selected unemployment beneficiaries.

• If the mayors’ assistants are correct, what is the chance that more than eight of the individuals investigated have jobs?

• If the mayors’ assistants are correct, what is the probability that only three of the individuals investigated have jobs?

• n=10

• p=0.40

• q=1-p= 0.60

^• Binomial distribution= ⁿc_x p^x q^n-x

• P(more than 8 have jobs)=[p(x=9) +p(x=10)]

=0.0016 + 0.0001= 0.0017

P(3 have jobs)=¹⁰c₃ (0.40)³ (0.60)^{7 =} 0.2150

Example 10

• A month later, Mayor Bruns picks up the morning edition of the city’s leading newspaper, the Times of India , and reads an expose of unemployment fraud. In this article, the newspaper claims that out of every 15 unemployment beneficiaries, the chance that four or more have jobs is 0.9095 and the expected number of employed beneficiaries exceeds 7. you are a special assistant to Mayor Bruns, who must respond to these claims at an afternoon press conference. She asks you to find the answers to the following two questions

• Are the claims of the Times of India consistent with each other?

• Does the first claim conflict with the opinion of mayor’s assistants?

• Binomial

• n=15

• p=?

• Claim 1

• p(x > 4)=0.9095 àp=0.40

• Claim 2

• np>p àp >7/15 = 0.467

• The claims are not consistent

• No. the assistants claim that p=0.40

Example 11

• R.V.Poppin, the concession stand manager for thee local hockey rink, just had 2 cancellations on his crew. This means that if more than 72,000 people come to tonight’s hockey game, the lines for hot dogs will constitute a disgrace to Mr. Poppin and will harm business at future games. Mr. Poppin knows from experience that the average number of people who come to the game is 67000 with a deviation of 4000 people.

• What is the future of Mr. Poppin’s business?

• Suppose Mr. Poppin can hire two temporary employees to make sure business won’t be harmed in the future at an additional cost of $200. If he believes the future harm to business of having more than 72,000 fans at the game would be $5,000. There will be no harm if 72,000 or fewer fans show up, and that the harm due to too many fans doesn’t depend on how many more than 72,000 show up. Should he hire the employees?. Explain.

• P(x>72000)=P(z>1.25)=0.5-.3944=.1056

• Loss to the future business=$528