Question

Suppose x has a distribution with μ = 21 and σ = 15.

(a) If a random sample of size n = 37 is drawn, find μ_x, σ_x and P(21 ≤ x ≤ 23). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(21 ≤ x ≤ 23) =

(b) If a random sample of size n = 57 is drawn, find μ_x, σ_x and P(21 ≤ x ≤ 23). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(21 ≤ x ≤ 23) =

(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is  ---Select--- larger than the same as smaller than part (a) because of the  ---Select--- same smaller larger sample size. Therefore, the distribution about μ_x is  ---Select--- the same wider narrower .

Answer 1

Let us assume that X is normally distributed with mean 21 and standard deviation 15.

We make this assumption considering that when we are given no underlying distribution it is observed that the population data follows normal distribution for a large size.

a) We are given that the random sample size 37. Using the normal distribution assumption one can obtain the value of the mean and variance of the sample by determining the expected value and the variance of the mean of the sample i.e. xbar.

Therefore:

or the mean is 21.

The standard error sigma x is given as:

which is equal to 2.47

The required probability using the normality of the sample mean can be written as:

P(21<=X<=23) = P(X<23) - P(X<21)

= P(Z<(23-21)/2.47) - P(Z<(21-21)/2.47)

= 0.2913

b) For a sample size of 57 we use the same procedure as before and get:

mux = 21

sigmax = 1.99

P(21<=X<=23) = 0.3429

c) The probability of part b is higher than that of part a since the standard deviation in part b is higher than that of part a. Due to higher standard deviation the distribution of X in part b is more peaked with thinner tails as compared to the distribution of x in part a. This difference in the two normal curves gives rise to a greater probability at center for part b as compared to part a.

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