Student Learning Outcome: The student will examine properties of the Central Lim
ID: 2907409 • Letter: S
Question
Student Learning Outcome: 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 of computer. Work will be graded for neatness and completeness The lifetime of a certain kind of battery is exponentially distributed, with an average of 40 hours. 1. We are interested in the lifetime of one batterv. 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 35 and 40 hours 4. Draw a graph to represent the probability in 3. Shade an appropriate region.Explanation / Answer
Let X = amount of time (in hours).
The time is known to have an exponential distribution with the average amount of time equal to forty hours. X is a continuous random variable since time is measured. It is given that
mu = 40 hours.
To do any calculations, you must know m, the decay parameter.
m = 1 / mu .
Therefore, m = 1 / 40 = 0.025.
The standard deviation, ?, is the same as the mean.
mu = ?
The distribution notation is X ? Exp(m). Therefore, X ? Exp(0.025).
The probability density function is
f(x) = me-mx. The number e = 2.71828182846.
The curve is: f(x) = 0.025e -0.025x where x is at least zero and m = 0.025
Find P(35 < x < 40).
The cumulative distribution function (CDF) gives the area to the left.
P(x < x) = 1 - e -mx
P(x < 40) = 1 - e(-0.025)(40)
= 0.6321 and
P(x < 35) = 1 - e (-0.025)(35)
= 0.5831
The probability is given by,
P(35 < x < 40)
= P(x < 40) - P(x <35)
= 0.6321 ? 0.5831 = 0.049.
75th percentile
P(x < k) = 0.75 and P(x < k) = 1 -e -0.025k
Therefore, 0.75 = 1 ? e ?0.025k and
e ?0.025k = 1 ? 0.75 = 0.25
Take natural logs: ln(e -0.025k ) = ln(0.25). So, -0.025k = ln(0.25)
Solve for k:k = ln(0.25) / (?0.025) = 55.45 hours.