Statistics and Probability Problems with Solutions
sample 2
Problems on statistics and probability are presented. The answers to these problems are at the bottom of the page.
Problems

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.
Solutions to the above Problems

 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)


If there 12 men, then there are 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 _{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


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)


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%
More References and linkselementary statistics and probabilities.
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