2. OBJECTIVES:
• Finds the mean and variance of the sampling distribution of the
sample mean. M11/12SP-IIId-5 5.
• Defines the sampling distribution of the sample mean for normal
population when the variance is: (a) known (b) unknown
M11/12SP-IIIe-1 6.
• Illustrates the Central Limit Theorem. M11/12SP-IIIe-2 7.
• Defines the sampling distribution of the sample mean using the
Central Limit Theorem. M11/12SP-III-3 8.
• Solves problems involving sampling distributions of the sample mean.
M11SP-IIIe-f-1
3. Suppose that all possible samples with size n are drawn from a
population with mean μ and standard deviation σ. Then, the
means obtained from all these possible samples will make up
a sampling distribution of the sample means exhibiting the
following properties:
1. The mean of the sample means is the same as the
population mean μ.
2. The standard error of the sample mean σ is equal to the
population standard deviation δ divided by the square root
of the sample size, thus, σ =
σ
𝑛
.
5. EXAMPLE 1:
• Random samples with size 4 are drawn from the population
containing the values 14, 19, 26, 31, 48, and 53. Find the mean
and standard error of the sample means.
•Solution:
μ =
14+19+26+31+48+53
6
μ =
191
6
μ = 31.83
6. •To get the standard error, compute first the population
standard deviation:
•Thus, the standard error of the mean is
7. EXAMPLE 2:
•A school has 900 junior high school
students. The average height of these
students is 68 in with a standard deviation
of 6 in. Suppose you draw a random sample
of 50 students. Find the mean, standard
deviation, and a variance of the distribution
of all sample means that can be derived
from the samples.
8. Solution:
a. The mean of the distribution of the sample means is the same as
the population mean. Thus
b. The standard error (or standard deviation) of the sample means is
equal to
c. The variance is the square of the standard deviation.
9. PRACTICE SET:
• Compute the mean, variance, and standard error of the sampling
distribution of the sample means given the sample size n taken
from the population and with mean μ and standard deviation σ.
1. n= 15 , μ = 18 , σ = 3
2. n= 8, μ = 24 , σ = 8
3. n= 35, μ = 54 , σ = 9
4. n= 20, μ = 3.25 , σ = 0.72
5. n= 30, μ = 8.12 , σ = 2.7
11. •Suppose a random variable X is given and
the population distribution of X is known
to be normal with mean μ and the
variance σ2. Then, it follows that the
sampling distribution of the mean of all
samples of size n selected from X is
normal with mean μ = and the
variance
12. • Knowing the normality of the sampling distribution of the
sample mean allows you to compute probabilities involving
means of samples using the standard normal distribution.
Any score from a normal distribution can be converted into
its equivalent standard normal score using the formula
• Now, since in a sampling distribution of the sample mean,
the random variable gives the statistic x̅ with mean μ and
standard deviation , the formula is derived as
13. EXAMPLE 1:
•Suppose that the systolic blood pressures of
a certain population are normally
distributed with mean μ = 125 and σ = 8 and
a sample size 36 is taken from this
population. What is the probability that a
single sample will have a mean blood
pressure of less than 122?
14. SOLUTION:
Given: μ = 125, σ = 8, x̅ = 122, n = 36
Substitute to the formula:
z =
122 −125
8
36
z =
−18
8
z =
−3
8
6
z = -2.25
In the z-table, P(-2.25) = P(2.25) = 0.4878
15. Continuation:
Thus,
P(x̅ < 122) = 0.5 – 0.4878
= 0.0122 or 1.22%
Conclusion: This means that the probability that a randomly
chosen sample from the population will have mean systolic
blood pressure less than 122 is 1.22%.
16. EXAMPLE 2:
• The average time that a person spends in a museum
follows a normal distribution with μ = 88 minutes and
a variance of 16 minutes. These parameters are taken
from a survey conducted to the visitors of the
museum in a particular day. Suppose a sample
consisting of 9 visitors is selected, what is the
probability that the average time these visitors spent
in the museum is between 87 and 92 minutes?
17. SOLUTION:
Given: μ = 88, σ = 16, x̅ = 87 and 92, n = 9
Substitute to the formula:
P(87 < x̅ < 92) = P(
87 −88
4
3
< Z <
92 −88
4
3
)
= P(-0.75 < Z < 3)
= P(0.75) + P(3)
= 0.2734 + 0.4987
P(87 < x̅ < 92) = 0.7721 or 77.21%
The probability that the mean time of 9 visitors falls between 87 and
92min is 77.21%.
19. CENTRAL LIMIT THEOREM
• Recall that when samples with any size n are taken from a
normally distributed population, then the distribution of all
sample means also follows a normal distribution. But what if
the population from which samples are taken is not normally
distributed? Can you conclude that the sampling distribution
of the sample mean will be normal? The answer to this
question is “YES” provided that the sample size is large
enough. This can be explained using the CENTRAL LIMIT
THEOREM.
20. CENTRAL LIMIT THEOREM
•Given a random variable X with mean μ and
variance σ², then, regardless whether the
population distribution of X is normally
distributed or not, as the sample size n gets
larger, the shape of the distribution of the
sample means taken from the population
approaches a normal distribution, with mean μ
and standard deviation
σ
𝑛
.
21. CENTRAL LIMIT THEOREM
•Note that the emphasis of the central limit
theorem is on the phrase “as the sample size
gets larger”. You can only assume the sampling
distribution to be normal when the sample is
large enough. Mostly, n ≥ 30 is sufficient to
approximate the normality of the sampling
distribution of the sample mean.
22. EXAMPLE 1:
•The average age of teachers in a certain
town is 34 with a standard deviation of
4. If the principal of ABC College
employs 100 teachers, find the
probability that the average age of these
teachers is less than 35.
23. SOLUTION:
• It is not given that the population is normally distributed but
since n > 30, then you can assume that the sampling
distribution of the mean ages of 100 teachers is normal
according to the CLT.
z =
35−34
4
100
z =
10
4
z =
1
4
10
z = 2.5
24. SOLUTION:
• Then,
P(X < 35) = P(Z < 2.5)
P(X < 35) = 0.5 + P(2.5)
P(X < 35) = 0.5 + 0.4938
P(X < 35) = 0.9938 or 99.38%
Therefore, the probability that the average age of
the teachers is less than 35 is 99.38%.
25. EXAMPLE 2:
•A random sample with size 16 is
selected from a normally distributed
population with a mean of 16 and a
standard deviation of 3.2. What is the
probability that the mean of the sample
falls between 14 and 16?
26. SOLUTION:
Given: μ = 16, σ = 3.2, x̅ = 14 and 16, n = 16
Substitute to the formula:
P(14 < x̅ < 16) = P(
14 −16
3.2
4
< Z <
16 −16
3.2
4
)
= P(-2.5 < Z < 0 )
= P(2.5) + P(0)
= 0.4938 + 0
P(87 < x̅ < 92) = 0.4938 or 49.38%
The probability that the mean of the sample falls between 14 and 16 is
49.38%.
28. PRACTICE SET:
1. A population has a mean age of 34 with a standard
deviation of 10. What is the probability that the mean age
of the sample with size 9 falls between 32 and 35,
considering that the given population follows a normal
distribution?
2. The heights of the students in a certain college are
normally distributed with an average height of 6.2 ft and a
standard deviation of 0.8 ft. If a random sample of 36
students is chosen, what is the probability that their height
fall between 6.6ft and 6.7ft?
