Question

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The Environmental Protection Agency (EPA) is concerned about pollution caused by factories that burn sulfur-rich fuel. In order to decrease the impact on the environment, factory chimneys must be high enough to allow pollutants to dissipate over a larger area. Assume that the mean height of chimneys in these factories is 100 meters (an EPA acceptable height) with standard deviation 12 meters. Use the appropriate Excel function to calculate each of the following. (Note – Part (b) will be answered by-hand.) [1 point each]

(b) Write your answer for Part (b) directly on the output. Suppose that the heights of all individual chimneys in the population vary according to an unknown distribution. Suppose that samples of 40 chimneys will be selected and the mean height of each sample, ?̅, will be recorded. What will be the shape of the sampling distribution of the ?̅ values? How do we know this?

(c) Find the probability that the mean height for a sample of 40 chimneys is greater than 102 meters.

(d) Find the probability that the mean height for a sample of 40 chimneys is between 101 and 103 meters.

Answer 1

Concept: Central limit theorem (CLT)

CLT says that if we have a population with a mean and standard deviation and if we draw a sufficiently large and random sample, then the means of the sample will follow a normal distribution. And this is valid even if the population distribution is not normal. So, there are two conditions for CLT to hold true

1. Sample size should be sufficiently high (n>=30)
2. The sample should be simply randomly drawn

b)

Here in this case, n>30 and sample is a simple random sample, so both the condition of CLT holds true, and hence the sampling distribution of ?̅ follows a normal distribution

c)

Let X be the mean height for a sample of chimneys, we need to find P(X>102)

Here in this case, as we deduced that mean would follow the normal distribution with mean = 100,

and standard deviation = standard deviation of population/sqrt(sample size)

=    = 1.8974

So, X ~ Normal distribution ( mean = 100, sd = 1.8974)

P(x>102) = 1 - P(X<=102) = 1 - NORM.DIST(102,100,1.8974,1) = 1- 0.854 = 0.1459 (this is excel syntax)

You can also calculate by translating into Z value

and then looking up the standard normal table

P(X>102)=P[Z>(102-100)/1.8974]=P(Z>1.06)=0.1459 (using standard normal distribution table)

d)

To find the probability that the mean height for a sample of 40 chimneys is between 101 and 103 meters

P(101<X<103) = NORM.DIST(103,100,1.8974,1) - NORM.DIST(101,100,1.8974,1) = 0.2422 (using excel formula)
Similarly, we can calculate from standard normal table as well, where
z1=(101-100)/1.89=0.53, z2=(103-100)/1.89=1.59.
P(101<X<103)=P(X<=103) - P(X<=101) = 0.9430-0.700916=0.2422 (using standard normal distribution table)