In order to estimate the mean mu of a large population, a statistician conducts

Question

In order to estimate the mean mu of a large population, a statistician conducts a sampling experiment with sample size 64. He computes the sample mean and sample standard deviation of his sample to be 34 and 2, respectively. (a) Let X denote the sample mean. The probability that mu lies in the interval [33.25, 34.75] is equivalent to the probability that X lies in the interval [mu - a, mu + a]. Find a. (b) What does the Central Limit Theorem have to say about the probability distribution of X? Your answer should reference the type of distribution possessed by X, as well as its mean and standard deviation. Draw and label the approximate probability distribution of X. (c) Using your answers to parts (a) and (b), Shade the area under the approximate probability distribution for X that corresponds to the probability that mu lies in the interval [33.25, 34.75]. Then, estimate this probability without using your calculator. (d) Use your calculator to check your answer to part (c)

Explanation / Answer

a. Here UCL=34.75=mu+a

And LCL=33.25=mu-a

Hence 2a=34.75-33.25=1.5

So a=0.75

b.      The central limit theorem states that if you have a population with mean and standard deviation and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30). If the population is normal, then the theorem holds true even for samples smaller than 30. In fact, this also holds true even if the population is binomial, provided that min(np, n(1-p))> 5, where n is the sample size and p is the probability of success in the population. This means that we can use the normal probability model to quantify uncertainty when making inferences about a population mean based on the sample mean.

For the random samples we take from the population, we can compute the mean of the sample means:

muxbar=34

and the standard deviation of the sample means:

sdxbar=sd/sqrt(n)=2/8=1/4=0.25

c. Here we need to find P(33.25<xbar<34.75), as it is normally distributed we can convert x to z.

P(33.25-34/0.25<z<34.75-34/0.25)=P(-3<z<3)=P(0<z<3)-P(0<z<-3)=0.4987+0.4987=0.9974

d. Using calculator we get P(-3<z<3)=0.9973

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