Answer to Question #126306 in Statistics and Probability for Ramovha Takalani

Question

A survey finds the following probability distribution for the age of a rented car:
Age in years 0 – 1 1 – 2 2 – 3 3 – 4 4 – 5 5 – 6 6 – 7
Probability 0.10 0.26 0.28 0.20 0.11 0.04 0.01
a. Find the probability that a rented car is between 0 and 4 years old.
b. Find the mean
Find the variance

Answer 1

Table

$\def\arraystretch{1.6} \begin{array}{r|c:c:c:c:c:c:c} \text{age} & 0-1 & 1-2 & 2-3 & 3-4 & 4-5 & 5-6 & 6-7 \\ \hline \text{probability} & 0.10 & 0.26 & 0.28 & 0.20 & 0.11 & 0.04 & 0.01 \\ \end{array}$

a)

$P(0-4) = P(0-1)+P(1-2)+P(2-3)+P(3-4) = 0.84$

b)

Solution:

Mean calculates using midpoints of range. So age of $\alpha-(\alpha+1)$ means that random variable is $\alpha + 0.5$.

$M(X) = 0.5\cdot0.10 + 1.5\cdot 0.26 + 2.5\cdot0.28+3.5\cdot0.20+4.5\cdot0.11+5.5\cdot0.04+6.5\cdot0.01 = 2.62$

$M(X^2) = 0.25\cdot0.10 + 2.25\cdot 0.26 + 6.25\cdot0.28 + 12.25\cdot0.20 + 20.25\cdot0.11 + 30.25\cdot0.04 + 42.25\cdot0.01 = 8.67$

$D(X)=M(X^2) - (M(X))^2 = 8.67 - 6.8644 = 1.8056$

Answer:

Mean is 2.62

Variance is 1.8056