7. Given that a representative sample of size 50 yields a mean of 20 and varianc

Question

7. Given that a representative sample of size 50 yields a mean of 20 and variance of 9, answer the following. [Hint: Normal distribution and Central Limit Theorem]

7.1   If the population is normal, determine the probability a randomly selected observation from the population will be between 11.5 and 28.5

7.2   If the population is not normal, determine the probability that a randomly selected observation from the population will be between 11.5 and 28.5

7.3   If the population is not normal, determine the probability that the population’s mean will be between 18.5 and 20.5

7.4   Determine the sampling distribution’s mean, standard deviation, and distribution (shape)

Please explain how you came to your answer, and if you used any functions in a Ti-83 or 84 calculator.

Thank you.

Explanation / Answer

According to the question it is given that,

Sample size(n)=50

Sample mean(µ)=20

Sample Variance(2)=9 Standard Deviation=3

It is also given that the underlying random variable X follows a normal distribution.

7.1 We are required to find P(11.5<X<28.5)

Standardising the above expression we have

P[(11.5-20)/3<(X-µ)/<(28.5-20)/3]= P(-2.83<Z<2.83) = P(Z<2.83)-P(Z<-2.83)

Now we use the standard normal probability tables to get the value of the above probabilities,

= 0.9977-0.0023 = 0.9954

*Alternatively the above expression can also be written as P(Z<2.83)-P(Z<-2.83) = P(Z<2.83)-[1-P(Z<2.83)]=0.9977-[1-0.9977]= 0.9977-0.0023=0.9954.

7.2 If we consider that the population is not normal then we have to use the central limit theorem which states that If X1, X2, ……. Xn is a sequence of independent, identically distributed (iid) random variables with finite mean µ and finite (non-zero) variance 2 then the distribution of( X-µ)/(/n) approaches the standard normal distribution, N(0,1) , as n tends to large values . Typically when n>30 we can apply the central limit theorem.

So, considering the probability again we have P(11.5<X<28.5)

Standardising the above expression we have P[(11.5-20)/(3/50)<( X-µ)/(/n) <(28.5-20)/(3/50)]

P(-20.035<Z<20.035)= P(Z<20.035)-P(Z<-20.035) ~1-0=1

7.3 Again if we consider that the population is not normal but we are given a large enough sample so we can say that Z=( X-µ)/(/n) follows a standard normal i.e N(0,1) distribution because of the central limit theorem.

Now we are required to find P(18.5<X<20.5)

Standardising the above expression we have P[(18.5-20)/(3/50)<( X-µ)/(/n) <(20.5-20)/(3/50)]

P(-3.5355<Z<3.5355)= P(Z<3.5355)-P(Z<-3.5355) ~0.9998-0.002=0.9996

7.4 Owing to the central limit theorem we can say that the sampling distribution of the mean is ~N(µ, 2/n). For the given example it becomes N(20, 3/50). This distribution is normal and hence symmetrical. Thus, the Central Limit Theorem provides a large-sample approximate sampling distribution for sample mean without the need for any distributional assumptions about the population. So for large n the distribution of the sample mean is N(µ, 2/n).This result is often called the z result. It transpires that the above result gives the exact sampling distribution of sample mean for random samples from a normal population.

The sampling distribution of Sample variance S2 when sampling from a normal population, withmean µ and variance 2, is (n-1)S2/ 2 ~2n-1 i.e it follows a chi-square distribution with n-1 degrees of freedom.

Whereas the distribution of sample mean is normal and hence symmetrical, the distribution of sample variance is positively skewed especially so for small n but becoming symmetrical for large n.

These are general results and no special functions on any calculator was used to get the results.

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