The student will examine properties of the Central Limit Theorem for averages. Use correct units where appropriate (inches, years, etc). The axes of graphs should be drawn with a ruler or computer. Work will be graded for neatness, completeness and correctness.
The lifetime of a certain kind of battery is exponentially distributed, with an average lifetime of 20 hours.
1. We are interested in the lifetime of one battery. Define the random variable X in words.
2. Give the distribution of X using numbers, letters, and symbols as appropriate.
3. Find the probability that the lifetime of one battery is between 20 and 25 hours.
4. Draw a graph to represent the probability in (3). Shade an appropriate region.

Given Information:

Lifetime of battery is exponentially distributed

Average lifetime of a battery = 20 hours

Required Information:

Define the random variable X in words = ?

Give the distribution of X = ?

P(20 < x < 25) = ?

Explanation:

1. We are interested in the lifetime of one battery. Define the random variable X in words.

X is a continuous random variable

X = lifetime of a battery (in hours) has an exponential distribution with the average lifetime of 20 hours.

Mean = μ = 20 hours.

Decay rate = m =   1 /μ = 1/20 = 0.05

The standard deviation is same as mean = σ = μ = 20

2. Give the distribution of X using numbers, letters, and symbols as appropriate.

The distribution notation is X~e^m

Therefore, X~e^0.25

The probability density function is

f(x) = me^–mx

f(x) = 0.05e^–0.05x

The cumulative distribution function is

P(x < x) = 1 – e–mx

P(x < x) = 1 – e–0.05x

3. Find the probability that the lifetime of one battery is between 20 and 25 hours.

P(20 < x < 25) = P(x < 25) – P(x < 20)

P(x < 25) = 1 – e(–0.05)(25) = 0.713

P(x < 20) = 1 – e(–0.05)(20) = 0.632

P(20 < x < 25) = 0.713 − 0.632 = 0.081

P(20 < x < 25) = 8.1 %

4. Draw a graph to represent the probability in (3). Shade an appropriate region

Attached as image

step-by-step explanation:

a and d