Download The Central Limit Theorem

MAT1028 Laboratory Session 3
The Central Limit Theorem
Log onto R, as in the first session.
Example 1: The weight of luggage that passengers take onto aircraft A is distributed
with mean 20kg and standard deviation 3.5kg. The aircraft carries 75 passengers.
(a) What is the probability that the weight of the luggage exceeds 1550kg?
By the Central Limit Theorem, the total weight of the luggage can be modelled by the
N(75 × 20, 75 ×3.5² ) distribution, i.e. by the N(1500, 918.75) distribution. Let W~
N(1500, 918.75), then the required probability is P(W>1550).
Calculate this probability with the command:
> 1- pnorm(1550,1500,sqrt(918.75))
Now calculate the probabilities that:
(b) The weight of the luggage exceeds 1520kg;
(c) The weight of the luggage is between 1480kg and 1520kg.
Note: by the Central Limit Theorem the underlying distribution need not be normal.
Example 2: The weight of luggage that passengers take onto aircraft B is distributed
with mean 22kg and standard deviation 5kg. The aircraft carries 70 passengers.
Calculate the probabilities that:
(a) The weight of the luggage exceeds 1650kg;
(b) The weight of the luggage is between 1510kg and 1570kg.
(c) The weight of the luggage on B exceeds that on A. (Hint: obtain an approximate
distribution for the difference of the totals.)
Example 3: From experience it is known that yields of a chemical in a manufacturing
process have mean 18g and variance 9g², per batch of raw material.
(a) What is the probability that for 45 batches the mean yield is less than 17.5g?
By the Central Limit Theorem, the mean yield can be modelled by the N(18, 3²/45 )
distribution, i.e. by the N(18, 0.2) distribution. Let X~ N(18, 0.2), then the required
probability is P(X<17.5).
Calculate this probability with the command:
> pnorm(17.5,18,sqrt(0.2))
Now calculate the probabilities that:
(b) The mean weight of 60 batches is less than 17.5g.
(c) The mean weight of 40 batches is not between 17.5g and 18.5g.
The Normal Distribution
Example 4: Items C and D are manufactured. The time taken to manufacture C has a
normal distribution with mean 10minutes and variance 2minutes². The time taken to
manufacture D has a normal distribution with mean 14minutes and variance
3minutes².
(a) What is the probability that the time taken to manufacture both C and D exceeds
26 minutes?
The sum of two normal random variables is also a normal random variable. Therefore,
the distribution for the time to manufacture both items is N(24,5). Required
probability is given by 1- pnorm(26,24,sqrt(5)).
(b) What is the probability that the time taken to manufacture C exceeds that for D?
Hint: Use results on the difference of two normal random variables.
(c) 15 items of type C are manufactured. What is the probability that the to mean
manufacture time for the 15 items exceeds 11 minutes? Explain why a sample size of
15 is sufficient to answer this question.
Answers
Example 1: (a) 0.04951523; (b) 0.2546818; (c) 0.4906364
Example 2: (a) 0.00427549; (b) 0.5267105; (c) 0.9981555
Example 3: (a) 0.1317762; (b) 0.0983528; (c) 0.2918405
Example 4: (a) 0.1855467; (b) 0.03681914; (c) 0.003084950