Question

Suppose x has a distribution with μ = 29 and σ = 21.

(a) If a random sample of size n = 47 is drawn, find μ_x, σ_x and P(29 ≤ x ≤ 31). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(29 ≤ x ≤ 31) =

(b) If a random sample of size n = 70 is drawn, find μ_x, σ_x and P(29 ≤ x ≤ 31). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(29 ≤ x ≤ 31) =

(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--- smaller than the same as larger than part (a) because of the  ---Select--- larger same smaller sample size. Therefore, the distribution about μ_x is  ---Select--- the same narrower wider .

Answer 1

a)

= = 29

= / sqrt(n)

= 21 / sqrt(47)

= 3.06

Using central limit theorem,

P( < x) = P(Z < (x - ) / )

So,

P(29 <= X <= 31) = P(X <= 31) - P(X <= 29)

= P(Z <= (31 - 29) / 3.06) - P(Z <= (29 - 29) / 3.06)

= P(Z < 0.65) - P(Z < 0)

= 0.7422 - 0.5

= 0.2422

b)

= = 29

= / sqrt(n)

= 21 / sqrt(70)

= 2.51

Using central limit theorem,

P( < x) = P(Z < (x - ) / )

So,

P(29 <= X <= 31) = P(X <= 31) - P(X <= 29)

= P(Z <= (31 - 29) / 2.51) - P(Z <= (29 - 29) / 2.51)

= P(Z < 0.80) - P(Z < 0)

= 0.7881 - 0.5

= 0.2881

c)

The standard deviation in part b is smaller than the part a) . because of larger sample size.

Therefore, the distribution about   is narrower