5.3. Central Limit Theorem 347variance is constant; in particular, it is free ofp. Hence, we seek a transformation
g(p) such that
g′(p)=c
√
p(1−p),for some constantc. Integrating both sides and making the change-of-variables
z=p,dz=1/(2
√
p)dp,wehaveg(p)=c∫
1
√
p(1−p)dp=2c∫
1
√
1 −z^2dz=2carcsin(z)=2carcsin(√
p).Takingc=1/2, for the statisticg
(
X)
=arcsin(√
X)
,weobtain√
n[
arcsin(√
X)
−arcsin (√
p)]D
→N(
0 ,1
4)
.Several other such examples are given in the exercises.
EXERCISES5.3.1.LetXdenote the mean of a random sample of size 100 from a distribution
that isχ^2 (50). Compute an approximate value ofP(49<X<51).5.3.2. LetX denote the mean of a random sample of size 128 from a gamma
distribution withα=2andβ= 4. ApproximateP(7<X<9).
5.3.3.LetYbeb(72,^13 ). ApproximateP(22≤Y≤28).
5.3.4.Compute an approximate probability that the mean of a random sample of
size 15 from a distribution having pdff(x)=3x^2 , 0 <x<1, zero elsewhere, is
between^35 and^45.
5.3.5.LetYdenote the sum of the observations of a random sample of size 12 from
a distribution having pmfp(x)=^16 ,x=1, 2 , 3 , 4 , 5 ,6, zero elsewhere. Compute an
approximate value ofP(36≤Y≤48).
Hint:Since the event of interest isY =36, 37 ,...,48, rewrite the probability as
P(35. 5 <Y < 48 .5).
5.3.6.LetYbeb(400,^15 ). Compute an approximate value ofP(0. 25 <Y/400).5.3.7.IfYisb(100,^12 ), approximate the value ofP(Y= 50).
5.3.8.LetYbeb(n, 0 .55). Find the smallest value ofnsuch that (approximately)
P(Y/n >^12 )≥ 0 .95.