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Statistics exam questions (taken from real previous exams)
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In a certain country, license plate numbers for cars consist of two letters followed by a four-digit number, such as TM8035 or AC2749. Assume that there are 26 letters.
a) How many different plates are possible if the two letters and also the four digits are allowed to be the same?
b) How many different plates are possible if the two letters are allowed to be the same but the four digits all have to be different?
c) Continue with the situation of part b. Let A be the event that a randomly drawn plate has a four-digit number that is greater than 5500. Calculate P(A).
In a certain country, license plate numbers for cars consist of two letters followed by a four-digit...
Here: unordered, without replacement.
a) N(9 4) = 9!/4!*5! = 6*7*8*9/24 = 126
b) A = 'only odd digits' = (5 4) = 5 --> P(A) = N(A)/N = 5/126 = 0,0397
c) N = 9*8*7*6 = 9!/5! = 3024
For a sample of 500 people, randomly drawn from the population of UvT students, the continuous variable X = ‘amount (in milliliter (ml)) of peanut butter consumed by the person in the week before' is noted. As a result, the following frequency distribution is obtained:
value frequency
[0, 1] 220
(1, 2] 123
(2, 3] 104
(3, 4] 39
(4, 5] 8
(5, 7] 6
total 500
a. Note that this classified frequency distribution has unequal class widths. Determine the frequency density at 6.
b. Let F be the accompanying cumulative distribution function. Calculate F(3.2).
c. Calculate the mean of the classified frequency distribution.
d. Also calculate the accompanying variance.
For a sample of 500 people, randomly drawn from the population of UvT students, the continuous varia...
a) (39+8+6)/500 = 0.106
b) F(3.2) = F(3) = (220+134+104)/500 = 0.894
c) mean x = (220x1 + 123x2 + ... + 6x6)/500 = 2.02
d) variance = (1/499)x{220x1²+ 123x2² + ... + 6x6²- 500x(2.02)²} = 1.2982
In a certain random experiment, the events A and B are considered. It is known that:
P(A)=0.55 and P(B)=0.15
Recall that P(A∪B)=P(A)+P(B)-P(A∩B). Below, always explain your answer. Calculate P(A∪B) if
a) it is additionally given that A and B are disjoint;
b) it is additionally given that B is a subset of A;
c) it is additionally given that A and B are independent;
d) it is additionally given that P(A│B)=0.8.
In a certain random experiment, the events A and B are considered. It is known that:
P(A)=0.55 a...
a) P(A∩B)= 0
So P(A∪B) = 0.55+0.15 = 0.7
b) B is a subset of A so:
A∩B = B and A∪B = A
P(A∪B) = P(A) = 0.55
c) A and B are independent
P(A∩B)=P(A)xP(B)
P(A∪B)= 0.55+0.15 - 0.55x0.15 = 0.6175
d) P(A│B) = 0.8 >> P(A∩B) = 0.8x0.15 = 0.12
P(A∪B) = 0.55 + 0.15 - 0.12 = 0.58
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