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:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

c. (3,2,6) is the answer.