Download Normal Approximations 3.6 Central Limit Theorem • Suppose X 1,X2

Normal Approximations
3.6 Central Limit Theorem
• Suppose X1, X2, . . . , Xn are independent
measurements coming from a population with
mean µ and variance σ 2. (i.e. the population does
not have to be normal)
• What is the sampling distribution of X?
The density function for X is approximately normal
with expected value µ and variance σ 2/n.
• n should be ‘large’. Larger n is needed when the
underlying population has more skewness, and
smaller n is okay for ‘light-tailed’ symmetric
populations. If the underlying is distribution is
normal, then any value of n is fine.
• Example. Under conditions of an accelerated life
test, the time to failure of a transistor may be
modelled as an exponential random variable with
expected value 200 hours.
◦ Suppose a random sample of 3 such
measurements is taken. What is the probability
that the average of the measurements exceeds
225 hours?
◦ Now consider a random sample of 36 such
measurements. What is the probability that the
average of the measurements exceeds 225
hours?
.
P (X > 225) = P (Z > .75) = .227
◦ Note that the standard deviation equals the
expected value for exponential random variables.
• Example. The lifetime of a type of battery is
normally distributed with expected value 10 hours
and standard deviation 3 hours. A package
contains 4 batteries. When camping, I plan to use
a flashlight that operates on 1 battery at a time
for 35 hours.
• Find the probability of running out of power early.
P (X̄ < 35/4) = P (Z < −.833) = .203
• A more expensive battery has the same expected
value, but a variance of 2.25 hours2. Find the
probability of running out of power early with this
brand.
P (X̄ < 35/4) = P (Z < −1.6) = .0548
• Example. The ACME elevator company uses cables
which will break when carrying more than 1000
pounds. 7 men board an elevator. If adult male
weight is normally distributed with expected value
150 pounds and standard deviation 15 pounds, find
the probability that the elevator cable will break.
P (X̄ > 1000/7) = P (Z > −1.26) = .896
• Example. The breaking strength of a rivet has an
expected value of 10000 psi and a variance of
250000 psi. The breaking strengths of 40 rivets are
measured. Find the probability that the average of
the measurements is between 9900 and 10200.
.
P (9900 < X̄ < 10200) =
P (−1.26 < Z < 2.53) = .890
• Find the probability that 1 of the rivets has a
breaking strength between 9900 and 10200?
We don’t know the distribution of the
measurements.
3.7 Approximations of Discrete Distributions
• We can approximate a binomial random variable
with large n by a normal random variable with
mean np and variance np(1 − p).
• We can approximate a Poisson random variable
with large λ by a normal random variable with
mean λ and variance λ.
Examples
• What is the approximate probability of observing
more than 80 tails (in our coin drop experiment) in
100 independent trials, if the true probability of a
tail is (a) 0.5? (b) 0.7?
• Chocolate chips are randomly distributed in cookies
with a mean of 15 per cookie. Find the
approximate probability that a cookie would have
more than 20 chocolate chips.
• Exercise Suppose a random sample of 1500 voters
is selected, and they are asked whether they
support the Conservative Party. If the proportion
supporting that party is .3 in the underlying
population, what is the approximate probability
that more than 33% of the sample says they
support the Conservative Party?