Statistics and Probability Problems with Solutions
Problems on statistics and probability are presented. The solutions to these problems are at the bottom of the page.
Given the data set
4 , 10 , 7 , 7 , 6 , 9 , 3 , 8 , 9
Find
a) the mode,
b) the median,
c) the mean,
d) the sample standard deviation.
e) If we replace the data value 6 in the data set above by 24, will the standard deviation increase, decrease or stay the same?
Find x and y so that the ordered data set has a mean of 42 and a median of 35.
17 , 22 , 26 , 29 , 34 , x , 42 , 67 , 70 , y
Given the data set
62 , 65 , 68 , 70 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101,
find
a) the median,
b) the first quartile,
c) the third quartile,
c) the interquartile range (IQR).
The exam grades of 7 students are given below.
70 , 66 , 72 , 96 , 46 , 90 , 50
Find
a) the mean
b) the sample standard deviation

Twenty four people had a blood test and the results are shown below.
A , B , B , AB , AB , B , O , O , AB , O , B , A
AB , A , O , O , AB , B , O , A , AB , O , B , A
a) Construct a frequency distribution for the data.
b) If a person is selected randomly from the group of twenty four people, what is the probability that his/her blood type is not O?
When a die is rolled and a coin (with Heads and Tails) is tossed, find the probability of obtaining
a) Tails and an even number,
b) a number greater 3,
c) Heads or an odd number,
A box contains red and green balls. The number of green balls is 1/3 the number of red balls. If a ball is taken randomly from the box, what is the probability that the ball is red?

The probability distribution of a random variable X is given in the table below
x P(X = x) 0 0.24 1 0.38 2 0.20 3 0.13 4 0.05
Find the mean μ and the standard deviation σ of X.
A committee of 6 people is to be formed from a group of 20 people. The committee has to have the number of women double that of the men. In how many ways can this committee be formed if there are 12 men?

Calculate the mean μ of the discrete data given in the frequency table below.x 1 2 3 4 5 6 f 2 6 10 6 2 2
A student's marks in five tests are 36%, 78%, 67%, 88% and 98%. The weights for the five tests are 1, 2, 2, 3, 3 respectively. Find the weighted mean μ of the five tests.
In a group of 40 people, 10 are healthy and every person the of the remaining 30 has either high blood pressure, a high level of cholesterol or both. If 15 have high blood pressure and 25 have high level of cholesterol,
a) how many people have high blood pressure and a high level of cholesterol?
If a person is selected randomly from this group, what is the probability that he/she
b) has high blood pressure (event A)?
c) has high level of cholesterol(event B)?
d) has high blood pressure and high level of cholesterol (event A and B)?
e) has either high blood pressure or high level of cholesterol (event A or B)?
f) Use the above to check the probability formula: P(A or B) = P(A) + P(B)  P(A and B).
A committee of 5 people is to be formed randomly from a group of 10 women and 6 men. Find the probability that the committee has
a) 3 women and 2 men.
a) 4 women and 1 men.
b) 5 women.
c) at least 3 women.
In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that
a) exactly 5 have access to the internet.
b) at least 6 students have access to the internet.
The grades of a group of 1000 students in an exam are normally distributed with a mean of 70 and a standard deviation of 10. A student from this group is selected randomly.
a) Find the probability that his/her grade is greater than 80.
b) Find the probability that his/her grade is less than 50.
c) Find the probability that his/her grade is between 50 and 80.
d) Approximately, how many students have grades greater than 80?

In a certain country last year a total of 500 million tons of trash was recycled. The chart below shows the distribution, in millions of tons, for the different types of trashes.
a) How many tons of Iron/Steel was recycled?
b) What percent of the recycled trash was glass?
Solutions to the above Problems

 The given data set has 2 modes: 7 and 9
 order data : 3 , 4 , 6 , 7 , 7 , 8 , 9 , 9 , 10 : median = 7
 (mean) : m = (3+4+6+7+7+8+9+9+10) / 9 = 7

x x  m (x  m)^{2} 4 3 9 10 3 9 7 0 0 7 0 0 6 1 1 9 2 4 3 4 16 8 1 1 9 2 4 sum = 44
sample standard deviation ≈ 2.35 (rounded to 2 decimal places)
 The standard deviation will increase since 24 is further from away from the other data values than 6.
x = 36 , y = 77

 median = 75
 first quartile = 69
 third quartile = 81
 interquartile range = 81  69 = 12
 median = 75

 mean = 70
 sample standard deviation ≈ 18.6 (rounded to 1 decimal place)
 mean = 70


class frequency A 5 B 6 AB 6 O 7
 1  (7/24) = 17/24 ≈ 0.71 (rounded to 2 decimal places)


 3/12 = 1/4
 6/12 = 1/2
 3/4

 3/4

 μ = Σ x P(X = x) = 0×0.24 + 1×0.38 + 2×0.20 + 3×0.13 + 4×0.05 = 1.37

1) using definition
σ = √[ Σ (x  μ)^{2} P(X = x) ]
= √[ (01.37)^{2}×0.24 + (11.37)^{2}×0.38 + (21.37)^{2}×0.2 + (31.37)^{2}×0.13 + (41.37)^{2}×0.05 ]
≈ 1.13 (rounded to 2 decimal places)
2) using computing formula
σ = √[ Σ x^{2} P(X = x)  μ^{2} ]
= √[ 0^{2}×0.24 + 1^{2}×0.38 + 2^{2}×0.2 + 3^{2}×0.13 + 4^{2}×0.05  1.37^{2}]
≈ 1.13 (rounded to 2 decimal places)
 μ = Σ x P(X = x) = 0×0.24 + 1×0.38 + 2×0.20 + 3×0.13 + 4×0.05 = 1.37


If there 12 men, then there are 20  12 = 8 women.
The committee has six people with the number of women double that of the men, hence the committee has 4 women and 2 men.
The number of ways of selecting 4 women from 8 is given by: _{8}C_{4} = 70.
The number of ways of selecting 2 men from 12 is given by: _{12}C_{2} = 66.
The number of selecting 4 women and 2 men to form the committee is given by :
_{8}C_{4} × _{12}C_{2} = 70 × 66 = 4620

If there 12 men, then there are 20  12 = 8 women.


μ = Σx_{i} × f_{i} / Σf_{i}
Σx_{i} × f_{i} = 1×2 + 2×6 + 3×10 + 4×6 + 5×2 + 6×2 = 90
Σf_{i} = 2 + 6 + 10 + 6 + 2 + 2 = 28
μ = 90 / 28 ≈ 3.21 (rounded to 2 decimal places)

μ = Σx_{i} × f_{i} / Σf_{i}


Let the marks be: x_{1} = 36%, x_{2} = 78%, x_{3} = 67%, x_{4} = 88%, x_{5} = 98% and the respective weights be: w_{1} = 1, w_{2} = 2, w_{3} = 2, w_{4} = 3, w_{5} = 3.
The weighted mean = Σ x_{i}×w_{i} / Σ w_{i}
Σ x_{i}×w_{i} = 36% × 1 + 78%×2 + 67%×2 + 88%×3 + 98%×3 = 884%
Σ w_{i} = 1 + 2 + 2 + 3 + 3 = 11
weighted mean = 884% / 11 = 80%

Let the marks be: x_{1} = 36%, x_{2} = 78%, x_{3} = 67%, x_{4} = 88%, x_{5} = 98% and the respective weights be: w_{1} = 1, w_{2} = 2, w_{3} = 2, w_{4} = 3, w_{5} = 3.


a) Let x be the number of people with both high blood pressure and high level of cholesterol. Hence (15  x) will be the number of people with high blood pressure ONLY and (25  x) will be the number of people with high level of cholesterol ONLY. We now express the fact that the total number of people with high blood pressure only, with high level of cholesterol only and with both is equal to 30.
(15  x) + (25  x) + x = 30
solve for x: x = 10
b) 15 have high blood pressure,hence P(A) = 15/40 = 0.375
c) 25 have high level of cholesterol, hence P(B) = 25/40 = 0.625
d) 10 have both,hence P(A and B) = 10/40 = 0.25
e) 30 have either, hence P(A or B) = 30/40 = 0.75
f) P(A) + P(B)  P(A and B) = 0.375 + 0.625  0.25 = 0.75 = P(A or B)



a) In what follows _{n}C_{r} = n! / [ (n  r)!r! ] and is the number of combinations of n objects taken r at the time and P(A) is the probability that even A happens.
There are _{16}C_{5} ways to select 5 people (committee members) out of a total of 16 people (men and women)
There are _{10}C_{3} ways to select 3 women out of 10.
There are _{6}C_{2} ways to select 2 men out of 6.
There are _{10}C_{3}*_{6}C_{2} ways to select 3 women out of 10 AND 2 men out of 6.
P(3 women AND 2 men) = _{10}C_{3}*_{6}C_{2} / _{16}C_{5} = 0.412087
b) similarly: P(4 women AND 1 men) = _{10}C_{4}*_{6}C_{1} / _{16}C_{5} = 0.288461
c) similarly: P(5 women ) = _{10}C_{5}*_{6}C_{0} / _{16}C_{5} = 0.0576923 (in _{6}C_{0} the 0 is for no men)
d) P(at least 3 women) = P(3 women or 4 women or 5 women)
since the events "3 women" , "4 women" and "5 women" are all mutually exclusive, then
P(at least 3 women) = P(3 women or 4 women or 5 women) = P(3 women) + P(4 women) + P(5 women)
≈ 0.412087 + 0.288461 + 0.0576923 ≈ 0.758240



a) If a pupil is selected at random and asked if he/she has an internet connection at home, the answer would be yes or no and therefore it is a binomial experiment. The probability of the student answering yes is 60% = 0.6. Let X be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that X = 5 is given by the binomial probability formula as follows:
P(X = 5) = _{8}C_{5} (0.6)^{5} (10.6)^{3} = 0.278691
b) P(X ≥ 6) = P(X = 6 or X = 7 or X = 8)
Since all the events X = 6, X = 7 and X = 8 are mutually exclusive, then
P(X ≥ 6) = P(X = 6) + P(x = 7) + P(X = 8)
= _{8}C_{6} (0.6)^{6} (10.6)^{2} + _{8}C_{7} (0.6)^{7} (10.6)^{1} + _{8}C_{8} (0.6)^{8} (10.6)^{0}
≈ 0.315394



a) x = 80 , z = (80  70)/10 = 1
Probablity for grade to be greater than 80 = 1  0.8413 = 0.1587
b) x = 50 , z = (50  70)/10 = 2
Probablity for grade to be less than 50 = 0.0228
c) The zscores for x = 50 and x = 80 have already been calculated above.
Probablity for grade to be between 50 and 80 = 0.8413  0.0228 = 0.8185
d) 0.1587 * 1000 ≈ 159 (rounded to the nearest unit)



a) 500  (170+90+60+50) = 130 tons of steel/iron was recycled.
b) 60/500 = 0.12 = 12% of the total recycled was glass.

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