TSTA401 Quantitative Methods for Accounting and Finance Assignment

This document presents a comprehensive solution to a quantitative methods assignment, likely for a finance or accounting course. The assignment covers key statistical concepts, including probability calculations using the z-table, applications of the normal distribution to real-world scenarios such as car fuel efficiency and child heights, and the calculation of confidence intervals for population means. The solution demonstrates step-by-step calculations for each question, providing clear explanations and interpretations of the results. The assignment includes problems involving finding probabilities for various events, determining confidence intervals, and applying statistical methods to analyze data related to real-world situations. The document showcases the application of statistical concepts and methods in finance and accounting.

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