Quantitative methods Student Name: Student Number: Date: 13th

Quantitative methods
Student Name:
Student Number:
Date: 13th February 2019

Question 1 [4 marks]
Part a) (2 marks)
Find the following probabilities by checking the z table
i) P(Z > -0.8)
Answer
P(Z > -0.8) = 1- .211855 = 0.788145
ii) P(-0.85<Z<0.45)
Answer
P(Z > -0.85) = 0.802348
P(Z < 0.45) = 0.673645
P(-0.85<Z<0.45) = 0.802348-0.673645 = 0.128703
iii) Z0.2
Answer
Z0.2 = 0.841481
Part b) (2 marks)
A new car has recently hit the market. The distance travelled on 1 gallon of fuel is normally
distributed with a mean of 65 miles and a standard deviation of 4 miles. Find the probability of
the following events.
i) The car travels more than 70 miles per gallon.
Answer
P(X > 70)
Z=70−65
4 = 5
4 =1.25
P(z > 1.25) = 0.10565
Thus the probability that the travels more than 70 miles per gallon is 0.10565
ii) The car travels less than 60 miles per gallon.
Answer
P(X < 60)

Z= 6 0−65
4 =−5
4 =−1.25
P(z < -1.25) = 0.10565
Thus the probability that the travels less than 60 miles per gallon is 0.10565
iii) The car travels between 55 and 70 miles per gallon.
Answer
P(55 < X < 70)
Z=55−65
4 =−10
4 =−2 . 5
P ( Z >−2.5 )=0.99379
Z=70−65
4 = 5
4 =1.25
P( z >1.25)=0.10565
P−2.5< Z 1.25¿=0.99379−0.10565=0.88814
Question 2 [4 marks]
Part a) (2 marks)
We assume that X is normal distributed with the mean value μ=800 and standard deviation
σ=100. Suppose the sample size n=16, find the following.
i) P( X >750)
Answer
Z=750−800
100 =−5 0
100 =−0 .5
P ( Z >−0 .5 ) =0.6914625
ii) P(750< X <1000)
Answer
Z=750−800
100 =−50
100 =−0.5
P ( Z >−0.5 )=0.6914625
Z=100 0−800
100 = 20 0
100 =2

P ( Z <2 )=0.97725
P ( −0.5<Z< 2 ) =0.97725−0.69146=0.28579
Part b) (2 marks)
The heights of children two-years-old are normally distributed with a mean of 32 inches and
standard deviation of 1.5 inches.
i) Find the probability that a two-year-old child is between 30 and 33 inches tall.
Answer
P(30 < X < 33)
Z=30−32
1.5 =−2
1.5 =−1.3333
P ( Z >−1.3333 ) =0.90879
Z=33−32
1.5 = 1
1.5 =0.6667
P ( Z <0.6667 ) =0.74752
P (−1.3333<Z <0.6667 )=0.90879−0.74752=0.16127
Thus the probability that a two-year-old child is between 30 and 33 inches tall is
0.16127
ii) If 9 children are randomly selected, find the probability that their mean heights
exceed 30 inches.
Answer
P(X>30)
Z=30−32
1.5/ √9 = −2
1.5/ 3 =−4.0
P ( Z >−4.0 )=0.99997
Thus if 9 children are randomly selected, find the probability that their mean heights
exceed 30 inches is 0.99997.
Question 3 [6 marks]
Part a) (3 marks)

Given the following information X =500, σ=12, n=50
Determine the 95% confidence interval estimate of population mean.
Answer
C . I : X ± zα/ 2 σ → 500 ±1.96∗12
500 ±23.52
Lower limit: 500−23.52=476.48
Upper limit: 500+23.52=523.52
Thus the 95% confidence interval estimate of population mean is between 476.48 and 423.52.
Part b) (3 marks)
A statistics practitioner calculated the mean and standard deviation from a sample of 51. They
are X =120 and s=15. Estimate the population mean with 95% confidence level
Answer
C . I : X ± zα/ 2
σ
√ n → 120 ± 1.96∗15
√ 51
12 0 ±1.96∗2.1004
120 ± 4.1168
Lower limit: 120−4.1168=115.8832
Upper limit: 120+ 4.1168=124.1168
Thus the 95% confidence interval estimate of population mean is between 115.8832 and
124.1168.
Question 4 [6 marks]
The number of pages printed before replacing the cartridge in a laser printer is normally
distributed with a mean of 11,500 pages and a standard deviation of 800 pages. A new cartridge
has just been installed.

Part a) (3 marks)
What is the probability that the printer produces more than 12,000 pages before this cartridge
must be replaced?
Answer
P(X > 12000)
Z=12000−11500
800 = 500
800 =0.625
P(z >0.625)=0.265986
Part b) (3 marks)
What is the probability that the printer produces fewer than 10,000 pages?
Answer
P(X < 10000)
Z=10 000−11500
800 =−−1500
800 =−1.875
P(z <−1.875)=0.030396