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