Fundamentals of Statistics and Probability | Solutions

Running head: FUNDAMENTALS OF STATISTICS AND PROBABILITY
FUNDAMENTALS OF STATISTICS AND PROBABILITY
Name of the Student
Name of the University
Author Note

FUNDAMENTALS OF STATISTICS AND PROBABILITY1
Question 1:
Total applicants = 400
Numbers of top 15% applicants = 400*0.15 = 60
Now, committee selected 12 students from the top 15%.
Now, let A = event that a person is admitted.
B = Person has the highest faculty rating
A ∩ B = event that a person is admitted and the person has highest faculty rating.
P(B) = 1/400
P(A ∩ B) = 1/(60C12*12C1)
Now, P(A|B) = P(A ∩ B)/P(B) = 400/(60C12*12C1) = 33.33/60C12
ii) Now, L = event that person has lowest faculty rating.
P(L) = 1/400
P(A ∩ L) = 0 (as committee always selected 12 students from the top 60 students)
Hence, P(A|L) = P(A ∩ L)/P(L) = 0
b) Type A defects = 3%
Type B defects = 2%
Type A and B = 0.4%
Hence, P(A∩B) = 0.004
P(A) = 0.03
P(B) = 0.02.

FUNDAMENTALS OF STATISTICS AND PROBABILITY2
Hence, P(the defect is type B given the product is known to have Type A defect)
= P(B|A) = P(A∩B)/P(A) = 0.004/0.03 = 0.1333 = 13.33%.
c)
Total sample data = 750
Dolomite samples = 480
Gamma reading more than 70 in dolomite = 50
Shale samples = 270
Gamma reading more than 70 in shale = 255
P(area should be mined given gamma reading more than 70) = P(dolomite | gamma reading
more than 70) + P(shale | gamma reading more than 70) = 50/750 + 255/750 = 0.4067 or
40.67%.
d) The CDF of exponential distribution function is
F(x;λ) = 1 – exp(-βx) x >= 0
= 0 x < 0
Here β = 1/λ
λ is the mean of exponential distribution
Given, car washing time of customers follows exponential distribution with mean 8 minutes.
Thus probability that a customer will take more than 11 minutes to complete job
= P(x>11) = 1 – P(x<11) = 1 – (1 - exp(-λx)) = exp(-(1/8)*11) = 0.2528 = 25.28%.
e) Probability that glass product has bubbles = p = 1/1000 = 0.001.

FUNDAMENTALS OF STATISTICS AND PROBABILITY3
i)
X = event of getting a product with bubbles.
n = 5000
p = 1/1000 = 0.001.
Now, as the x values are discrete hence binomial distribution can be applied for modelling to
exactly compute the probabilities with p = 0.001 and n = 5000.
Binomial pmf is given by,
f(x) = nCx * p^x * (1-p)^x
Hence, P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= 5000C0 * 0.001^0 * (1-0.001)^5000 + 5000C1 * 0.001^1 * (1-0.001)^4999 + 5000C2 *
0.001^2 * (1-0.001)^4998 + 5000C3 * 0.001^3 * (1-0.001)^3
= 0.2649 or 26.49%
ii) Now, an appropriate model is applied where the binomial distribution is approximated to
Poisson distribution with np = λ= 5000*0.001 = 5.
The pmf of Poisson distribution is
f ( x )= ( λx )∗exp (−λ )
x !
Hence, by the approximate model
P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= ( 50 )∗exp (−5 )
0 ! + ( 51 )∗exp ( −5 )
1! + ( 52 )∗exp (−5 )
2! + ( 53 )∗exp (−5 )
3!