MCB1007 Introduction to Probability and Statistics Fall 2016-2017
Final Exam January 3, 2017
No:
Name:
Section:
Justify your answers to get full credit − Exam Duration 90 Minutes 1) At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed
to be normally distributed with mean36 oz. and standard deviation 0.1 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of the bottle goes below 35.8 oz. or above 36.2 oz., then the bottle will be declared out of control.
(a) Find the probability that a bottle will be declared out of control.
Answer. The process is out of control if P (X < 35.8) or P (X > 36.2).
P (X < 35.8) + P (X > 36.2) = P
Z < 35.8 − 36 0.1
+ P
Z > 36.2 − 36 0.1
= P (Z < −2) + P (Z > 2)
= (0.5 − 0.4772) + (0.5 − 0.4772)
= 2 × 0.0228 = 0.0456.
1a / 9
1b / 6
2a / 8
2b / 10
3a i / 5
3a ii / 5
3b / 7
4a / 8
4b / 8
5a / 8
5b / 8
6a / 9
6b / 9
/100
(b) Find the probability that the number of bottles found out of control in an eight-hour day will be zero.
Answer. This is binomial with n = 16 θ = 0.0456.
P (X = 0) = b(0; 16, 0.0456) =
16 0
0.04560(1 − 0.0456)16= 0.954416= 0.4739.
2) (a) Based upon past experience, 40% of all customers at Miller’s Automotive Service Station pay for their purchases with a credit card. If a random sample of 600 customers is selected, what is the approximate probability that at least 250 pay with a credit card? (Hint: Use the Normal approximation) Answer. We use the Normal Distribution to approximate the Binomial with
μ = nθ = 600 × 0.40 = 240, σ =
nθ(1 − θ) =
600 × 0.40(1 − 0.40) =√
144 = 12:
P (X ≥ 250) = P (X > 249.5) ≈ P
Z > 249.5 − 240 12
= P (Z > 0.79)
≈ 0.5 − 0.2852 = 0.2148.
(b) A door to door salesman keeps a record of the number of homes he visits each day Homes visited 0 − 9 10 − 19 20 − 29 30 − 39 40 − 49
Frequency 3 8 24 60 21
Find the mean number, the median and the modal class of homes visited.
Answer.
Mean: ¯x = 4.5 × 3 + 14.5 × 8 + 24.5 × 24 + 34.5 × 60 + 44.5 × 21
116 = 1861
58 = 32.0862069.
Median: 29.5 +39.5 − 29.5
60 × 23.5 = 401
12 = 33.41¯6.
Modal Class: 30 − 39.
3) (a) Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
(i) Find the probability that at least two passes in a given minute.
Answer. μ = 30060 = 5 per minute.
P (x ≥ 2) = 1 − P (x < 2) = 1 −
e−550
0! +e−551 1!
= 0.9596.
(ii) What is the expected number passing in two minutes? Find the probability that this expected number actually pass through in a given two-minute period.
Answer.
E(X) = 5 × 2 = 10, P (X = 10) = e−101010
10! = 0.12511.
(b) Suppose an urn contains 4 red and 10 blue balls and that balls are drawn one after another from this urn until a red ball is obtained. What is the probability that exactly six balls are drawn?
Answer.
P (First red ball is sixth ball) = P (First five balls are blue) × P (Sixth ball is red)
=
10
5
4
0
14
5
× 4 9 = 18
143 ×4 9 = 8
143 = 0.056.
4) (a) Four different prizes are randomly put into boxes of cereal. One of the prizes is a free ticket to the cinema. Suppose that a family of four (called Family A) decides to buy this cereal until obtaining four free tickets to the cinema. What is the probability that the family will have to buy 10 boxes of cereal to obtain the four free tickets to the cinema?
Answer. P (Selecting ticket to the cinema from box) = 14 = 0.25. So, required probability is
9 3
(0.25)4(0.75)6 = 0.058399.
(b) Nine ping pong balls are labeled with the integers 1, 2, 3, 4, 5, 6, 7, 8 and 9 respectively. If three balls are selected at random from these nine, what is the probability that more even-numbered balls than odd-numbered balls are selected?
Answer. Even balls: 2,4,6,8 and Odd balls: 1,3,5,7,9
P (number of even balls > number of odd balls) =
4
2
5
1
+4
3
5
0
9
3
= 17
42 = 0.40476.
5) A university wants to estimate the average monthly lunch costs of students. A random sample of size 100 selected from all students with mean 150 TL and standard deviation 30 TL.
(a) Construct a 95% confidence interval for μ.
Answer. Forn = 100, ¯x = 150, σ = 30 and z0.025 = 1.96
150 − 1.96√30
100 < μ < 150 + 1.96√30 100
150 − 5.88 < μ < 150 + 5.88 ⇒ 144.12 < μ < 155.88.
(b) With what degree of confidence could we say that the mean of monthly lunch costs is 150 ± 2.
Answer.
150 − zα/2√30
100 < μ < 150 + zα/2√30
100 ⇒ 150 − 3zα/2 < μ < 150 + 3zα/2
3zα/2= 2 ⇒ zα/2= 2/3 = 0.¯6
Degree of confidence: 2 × 0.2486 = 0.4972 or 50%.
6) From the set of numbers {2, 4, 6} a random sample of size 2 will be selected. Determine the sampling distribution of the mean and find the mean and the standard deviation of the population and the sampling distribution of mean if sampling were done
(a) Without replacement.
Answer.
(x1, x2) (2,4) (2,6) (4,6)
¯x 3 4 5 ⇒ ¯x 3 4 5
f(¯x) 1/3 1/3 1/3 For the population distribution:
Mean=μX = 2+4+63 = 4, Standard Deviation=σX =
(2−4)2+(4−4)2+(6−4)2
3 = 2√36 = 1.632993162.
For the sampling distribution of the mean:
Mean=μX¯ = μX = 4, Standard Deviation=σX¯ = √σXn
N−nN−1 = 2
√6
√3
2
3−2
3−1 = √36 = 0.8164965809.
(b) With replacement.
Answer.
(x1, x2) (2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)
¯x 2 3 4 3 4 5 4 5 6
¯x 2 3 4 5 6
f(¯x) 1/9 2/9 3/9 2/9 1/9 For the population distribution:
Mean=μX = 4, Standard Deviation=σX = 2√36 = 1.632993162.
For the sampling distribution of the mean:
Mean=μX¯ = μX = 4, Standard Deviation=σX¯ = √σXn = 2
√6
√3
2 = 2√33 = 1.154700538.
