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6.5
What does the Central Limit Theorem tell us about the sampling distribution of the sample mean?
從一個母體中抽樣一組 n 筆資料，算出樣本平均 x ，如果 n 很大（沒有一定的標準，通常母
體為分配愈對稱時，所需樣本數愈少，通常要求 n  30）
，則 x 的分配會接近常態分配，而且 x 的
平均數仍為原母體平均數  ，但 x 的標準差（即標準誤）為

n
（  為原母體的標準差）。
6.9
Suppose that we will randomly select a sample of 64 measurements from a population having a mean
equal to 20 and a standard deviation equal to 4.
a. Describe the shape of the sampling distribution of the sample mean x Do we need to make any
assumptions about the shape of the population? Why or why not?
※ Central Limit Therorem, normal distribution
b. B .Find the mean and the standard deviation of the sampling distribution of the sample mean x .
※ x =20
4

※ standard deviation of x （standard error）  x 
=
 0.5
n
64
c. Calculate the probability that we will obtain a sample mean greater than 21; that is, calculate
P(.x > 21). Hint: Find the z value corresponding to 21 by using μ and σ because we wish to
calculate a probability about x. Then sketch the sampling distribution and the probability
x  0 21  20
 2 p(z>2)=0.0228
=
z

0 .5
n
d. Calculate the probability that we will obtain a sample mean less than 19.385 ; that is calculate
P(x.< 19.385).
x  0 19.385  20
 1.23 p(z>-1.23)=0.1093
=
z

0.5
n
6.16
Suppose that we will randomly select a sample of n units from a population and that we will compute
the sample proportion p of these units that fall into a category of interest. If we consider the sampling
distribution of p:
a. If the sample size n is large, the sampling distribution of p is approximately a normal distribution.
What condition must be satisfied to guarantee that n is large enough to say that p is normally
distributed?
np  5 and n(1  p )  5
b. Write formulas that express the central tendency and variability of the population of all possible
sample proportions. Explain what each of these formulas means in your own words.
 x  p, x 
p(1  p)
n
6.21
Suppose that we will randomly select a sample of n = 100 units from a population and that we will
compute the sample proportion p of these units that fall into a category of interest. If the true
population proportion p equals .9:
a. Describe the shape of the sampling distribution of p. Why can we validly describe the shape?
np  100  0.9  90  5 and n(1  p)  100  0.1  10  5
b. Find the mean and the standard deviation of the sampling distribution of p.
1
p(1  p)
0.9  0.1

 0.03
n
100
c. Calculate the following probabilities about the sample proportion p. In each case sketch the
sampling distribution and the probability.
(1) P( pˆ  0.96 )
(2) P(.855  p̂  .945)
(3) P( p̂ < .915.)

pˆ  0.96
p̂ < .915.
.855  p̂  .945
p  p0
z
z
<0.5
2
 1.5
p0 (1  p0 ) / n
p
0.0228
0.8664
0.6915
 x  p  0.9,  x 
6.33
Each day a manufacturing plant receives a large shipment of drums of Chemical ZX-900 These drums
are supposed to have a mean fill of 50 gallons, while the fills have a standard deviation known to be .6
gallon.
a. Suppose that the mean fill for the shipment is actually 50 gallons. If we draw a random sample of
100 drums from the shipment, what is the probability that the average fill for the 100 drums is
between 49.88 gallons and 50.12 gallons?
x  0
=  2 p(2  z  2)  09544
z

n
b. The plant manager is worried that the drums of Chemical ZX-900 are underfilled . Because of this.
she decides to draw a sample of 100 drums from each daily shipment and will reject the shipment
(send it back to the supplier) if the average fill for the 100 drums is less than 49.85 gallons.
Suppose that a shipment that actually has a mean fill of 50 gallons is received What is the
probability that this shipment will be rejected and sent back to the supplied
x  0
=  2.5 p( z  2.5)  0.0062
z

n
2
※ Z0.0228=2
95%C.I.=[50.6  2  1.62
※ Z0.0228=2, 95% C.I.=[50.6  2  1.62
5
]=[49.151,52.049]
40
3
]=[50.088,51.112]