Question

The number of tickets purchased by an individual for Beckham College’s holiday music festival is a uniformly distributed random variable ranging from 2 to 10. Find the mean and standard deviation of this random variable. (Round your answers to 2 decimal places.)

Answer 1

Answer:

The mean of the distribution is 4.00 and the standard deviation is 2.31

Explanation:

Given

Range = 2 to 10

Type of distribution = uniform distribution

Required

1. Mean

2. Standard Deviation

The mean of uniform distribution is calculated as thus;

$Mean = \frac{1}{2}(a + b)$

Where b and a are the intervals of the distribution

b = upper bound = 10

a = lower bound = 2

So,

$Mean = \frac{1}{2}(a + b)$

Substitute 10 for b and 2 for a

$Mean = \frac{1}{2}(2 + 10)$

$Mean = \frac{1}{2}(12)$

$Mean = \frac{1}{2} * 12$

$Mean = 6$

$Mean = 6.00$ (Approximated to 2 decimal places)

The standard deviation of uniform distribution is calculated as thus;

σ = √σ²

Where σ represents the standard deviation and σ² represents the variance.

Calculate variance

σ² = Var

$Var = \frac{(b-a)^2}{12}$

Substitute 10 for b and 2 for a

$Var = \frac{(10 - 2)^2}{12}$

$Var = \frac{8^2}{12}$

$Var = \frac{64}{12}$

$Var = 5.33$

Recall that

σ = √σ² = √Var

Substitute 5.33 for Var

σ = √5.33

σ = 2.309401076758503

σ = 2.31 (Approximated)

Hence, the mean of the distribution is 4.00 and the standard deviation is 2.31