Chapter 5

Fundamental Sampling
Distributions
 Chapter 5
Fundamental Sampling Distributions
 Random Sampling
 Some Important Statistics
 Sampling Distribution
 Central Limit Theorem
Random Sampling
The outcome of a statistical experiment may be recorded
either as a numerical value or as a descriptive
representation.
Definition
A population consists of the totality of the observations
with which we are concerned. (whether their number be
finite or infinite).
Definition
A sample is a subset of a population.
Definition
Any function of the random variables constituting a
random sample is called a statistic.
Some Important Statistics
Location Measures of a Sample:
let 𝑋1 , 𝑋2 ,…., 𝑋𝑛 represent 𝑛 random variables
1. Mean
𝑛
1
𝑋ത = ෍ 𝑋𝑖
𝑛
𝑖=1
2. Median
𝑥𝑛+1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
,
2
𝑋ෘ = 1
(𝑥𝑛 + 𝑥𝑛+1 ), 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
2 2 2
3. Mode is the value of the sample that occurs most
often.
Some Important Statistics
Variability Measures of a Sample:
let 𝑋1 , 𝑋2 ,…., 𝑋𝑛 represent 𝑛 random variables
1. Variance
𝑛
1
2
𝑆 = ෍(𝑋𝑖 − 𝑋) ത 2
𝑛−1
𝑖=1
2. Standard Deviation
𝑆 = 𝑆2
3. Range
𝑅 = 𝑋𝑚𝑎𝑥 − 𝑋𝑚𝑖𝑛
EXAMPLE 1
The lengths of time, in minutes, that 10 patients waited
in a doctor’s office before receiving treatment were
recorded as follows: 5, 11, 9, 5, 9, 15, 6, 10, 5, and 10.
Treating the data as a random sample, find
(a) the mean;
(b) the median;
(c) the mode.
(d)the range;
(e) The variance
(f) the standard deviation
Solution
𝑛 = 10 (even), σ𝑛𝑖=1 𝑋𝑖 =85
85
(a) mean 𝑥ҧ = =8.5
10
(b) Rearrange 5,5,5,6,9,9,10,10,11,15
1 9+9
median = (𝑥5 + 𝑥6 )= =9
2 2
(c) mode is: 5
(d) range=15-5=10
1 2 96.5
(e) variance 𝑠 =2 σ𝑛𝑖=1(𝑋𝑖 − 8.5) = =10.72
10−1 9
(f) the standard deviation is s=3.27
 Sampling Distribution
Inference about the Population from Sample Information.
Definition
The probability distribution of a statistic is called a
sampling distribution.
For Example:
 sampling distribution of Mean
 Sampling Distribution of the Difference between Two
Means
 sampling distribution of Proportion
 sampling distribution of Variance
Sampling Distribution Of Means And The Central
Limit Theorem
Suppose that a random sample of n observations is
taken from a normal population with mean μ and
variance σ2 .
1
ഥ
Hence, 𝑋 = (𝑋1 +𝑋2 +…+𝑋𝑛 ) has a normal distribution
𝑛
with mean
1
𝜇 ഥ𝑋 = (𝜇+ 𝜇+….+ 𝜇)= 𝜇 and variance
𝑛
n times
1 σ2
𝜎2 𝑋 =𝑛2 (σ +σ +…+σ )= 𝑛
2 2 2
ഥ
Central Limit Theorem
If 𝑋ത is the mean of a random sample of size n taken
from a population with mean μ and finite variance σ2 ,
then the limiting form of the distribution of
𝑋ത − 𝜇
𝑍=
𝜎/ 𝑛
as n→∞, is the standard normal distribution n(z; 0, 1).
EXAMPLE 2
An electrical firm manufactures light bulbs that have a
length of life that is approximately normally distributed,
with mean equal to 800 hours and a standard deviation
of 40 hours. Find the probability that a random sample
of 16 bulbs will have an average life of less than 775
hours.
Solution:
𝑛=16, 𝜇=800, 𝜎=40
𝑋ത − 𝜇 775 − 800 −5
𝑍= = = = −2.5
𝜎/ 𝑛 40/ 16 10
𝑃 𝑋ത < 775 = 𝑃 𝑍 < −2.5 = 1 − 𝑃 𝑍 < 2.5 =0.0062
EXAMPLE 3
A tobacco company claims that the amount of nicotine
in its cigarettes is a random variable with mean 2.2 mg
and a standard deviation 0.5 mg. What is the probability
that the sum of nicotine content in a sample of 100
cigarettes would have been as higher than 230 mg?
Solution
230
P(σ𝑛𝑖=1 𝑋𝑖 ത
> 230) = 𝑃 𝑋 > = 𝑃(𝑋ത > 2.3)
100
2.3−2.2
𝑃(𝑍 > ) =𝑃 𝑍 > 2 = 1 − 0.9772 = 0.0228
0.5/10
EXAMPLE 4
A random sample of size 169 is taken from normal
population with variance 400. What is the probability
that the absolute difference between the sample mean
and the population mean is greater than 4
Solution
P 𝑋ത − 𝜇 > 4 = 1 − P 𝑋ത − 𝜇 ≤ 4
= 1 − P −4 ≤ 𝑋ത − 𝜇 ≤ 4
= 1 − 𝑃 −2.6 ≤ 𝑍 ≤ 2.6
= 1 − {𝑃 𝑍 ≤ 2.6 − 𝑃 𝑍 ≥ 2.6 }
= 1 − 𝑃 𝑍 ≤ 2.6 + 1 − 𝑃 𝑍 ≤ 2.6
= 0.0094
Sampling Distribution of the Difference
between two Means
If independent samples of size 𝑛1 and 𝑛2 are drawn at
random from two populations, discrete or continuous,
with means 𝜇1 and 𝜇2 and variances 𝜎1 2 and 𝜎2 2
respectively, then the sampling distribution of the
differences of means, 𝑋ത1 − 𝑋ത2 is approximately normally
distributed with mean and variance given by
𝜎1 2 𝜎2 2
𝜇𝑋ത1−𝑋ത2 = 𝜇1 − 𝜇2 and 𝜎 2𝑋ത1−𝑋ത2 = +
𝑛1 𝑛2
Hence,
(𝑋ത1 − 𝑋ത2 ) − (𝜇1 − 𝜇2 )
𝑍=
𝜎1 2 𝜎2 2
+
𝑛1 𝑛2
is approximately a standard normal variable.
EXAMPLE 5
The television picture tubes of manufacturer A have a
mean lifetime of 6.5 years and a standard deviation of
0.9 year, while those of manufacturer B have a mean
lifetime of 6.0 years and a standard deviation of 0.8
year. What is the probability that a random sample of 36
tubes from manufacturer A will have a mean lifetime
that is at least 1 year more than the mean lifetime of a
sample of 49 tubes from manufacturer B?
Solution: Manufacturer A : Population 1
Manufacturer B : Population 2
Population 1 Population 2
𝑛1 =36 𝑛2 = 49
𝜇1 =6.5 𝜇2 = 6
𝜎1 =0.9 𝜎2 = 0.8
ത ത
(𝑋1 −𝑋2 )−(𝜇1 − 𝜇2 )
ത ത
P(𝑋1 − 𝑋2 ≥ 1.0)= P( ≥ 1.0)
2 𝜎1 2 𝜎
+ 2
𝑛1 𝑛2
1.0−(6.5−6)
= P(𝑍 ≥ )
0.81 0.64
36
+ 49
1−0.5
=P(Z ≥ )=1- P(Z ≤2.65) from table
0.189
=1-0.9960=0.0040
 EXAMPLE 6
Two independent experiments are run in which two
different types of paint are compared. Eighteen
specimens are painted using type A, and the drying time,
in hours, is recorded for each. The same is done with type
B. The population standard deviations are both known to
be 1.0. Assuming that the mean drying time is equal for
the two types of paint, find P(𝑋ത𝐴 − 𝑋ത𝐵 > 1.0), where
𝑋ത𝐴 and𝑋ത𝐵 are average drying times?
Solution

(𝑋ത 𝐴 −𝑋ത 𝐵 )−(𝜇𝐴 − 𝜇𝐵 ) 1−0 1

ത ത
(𝑋𝐴 −𝑋𝐵 )−(𝜇𝐴 − 𝜇𝐵 )
ത ത
P(𝑋𝐴 − 𝑋𝐵 > 1.0)=P( >3)
𝜎𝐴 2 𝜎𝐵 2
+
𝑛𝐴 𝑛𝐵

∴P(Z>3)=1-P(Z<3)
=1-0.9987
=0.0013
Sampling Distribution of Proportion
Suppose that a random sample of n observations is taken
from a normal population with mean μ and variance σ2 .
Hence, ෡𝑃 has a normal distribution with mean
𝑋 𝑛𝑝
෠
𝜇𝑃෠ = 𝐸 𝑃 = E = =𝑝
𝑛 𝑛
and variance
𝜎2𝑋 𝑛𝑝𝑞 𝑝𝑞
𝜎2 𝑃෠ = 𝜎2 𝑋/𝑛 = 𝑛2 =
𝑛2
=
𝑛
Then, the sampling distribution of proportion is
෡𝑃 − 𝑝
𝑍=
𝑝𝑞/𝑛
EXAMPLE 7
Suppose that in a certain human population 0.08 are
colorblind. If a random sample of 150 individuals from
this population is selected. What is the probability that
the proportion in the sample who are colorblind will be
greater than 0.15?
Solution
𝑝 = 0.08 , 𝑞 = 1 − 0.08 = 0.92 , 𝑛 = 150
෡𝑃 − 𝑝 0.15 − 0.08
𝑃 𝑝Ƹ > 0.15 = P Z > = 𝑃(𝑍 > )
𝑝𝑞 (0.08)(0.92)
𝑛 150
𝑃 𝑍 > 3.1 = 1 − 𝑃 𝑍 < 3.1 = 1 − 0.999 = 0.001