Lesson 5.1.-_- Finding Variance and Standard Deviation of sampling distribution.pptx

Lesson 5.1: FINDING THE
MEAN AND THE VARIANCE OF
THE SAMPLING DISTRIBUTION
OF THE SAMPLE MEANS
ANGELICA C. IGNI
SUBJECT TEACHER
UNIT 5: FINDING THE MEAN AND THE VARIANCE
OF THE SAMPLING DISTRIBUTION OF THE SAMPLE
MEANS

Example: The following table gives the sum of
tutorial rate of six teachers in Central Luzon per
month. Suppose that random samples of size 4 are
taken from this population of six teachers, do the
following tasks.
TEACHER
Tutorial Rate (in
thousand pesos) X
A 8
B 12
C 16
D 20
E 24

1) Solve for the mean of the population .
2) Solve for the mean of the sampling distribution of the
sample means .
3) Compare and
4) Solve for the variance ( ) and the standard deviation
(σ) of the population.
5) Solve the variance () and the standard deviation ( ) of
the sampling distribution of the sample means .
6) Compare σ and .
STEPS FINDING THE MEAN AND THE
VARIANCE OF THE SAMPLING
DISTRIBUTION OF THE SAMPLE MEANS

SOLUTIONS:
1) The population mean is solved as follows:
𝝁 = ¿
8+ 12+ 16+20 +24 +28
6
¿
1 08
6
¿ 18
Therefore, the population
mean of the tutorial rates of the
select teachers in Central Luzon
is 18 thousand pesos per
month.

2) To solve for the mean of the sampling distribution of the
sample means, the following steps are to be considered.
a) Identify the possible samples of size 4 and compute
their individual means
Possible Sample Sample Mean ()
8, 12, 16, 20
8, 12, 16, 24
8, 12, 16, 28
8, 12, 20, 24
8, 12, 20, 28
8, 12, 24, 28
8, 16, 20, 24
8, 12, 16, 20 14
8, 12, 16, 24 15
8, 12, 16, 28 16
8, 12, 20, 24 16
8, 12, 20, 28 17
8, 12, 24, 28 18
8, 16, 20, 24 17

2) To solve for the mean of the sampling distribution of the
sample means, the following steps are to be considered.
a) Identify the possible samples of size 4 and compute
their individual means
8, 16, 24, 28
8, 20, 24, 28
12, 16, 20, 24
12, 16, 20, 28
12, 16, 24, 28
12, 20, 24, 28
16, 20, 24, 28
8, 16, 24, 28 19
8, 20, 24, 28 20
12, 16, 20, 24 18
12, 16, 20, 28 19
12, 16, 24, 28 20
12, 20, 24, 28 21
16, 20, 24, 28 22

b) Construct the sampling distribution table for the
sample means and multiply the sample means to
their probabilities.
Sample Mean Frequency (F) Probability P() P()
14
15
16
17
18
19
20
21
22
14 1
15 1
16 2
17 2
18 3
19 2
20 2
21 1
22 1
14 1 1/15
15 1 1/15
16 2 2/15
17 2 2/15
18 3 3/15
19 2 2/15
20 2 2/15
21 1 1/15
22 1 1/15
14 1 1/15 14/15
15 1 1/15 15/15
16 2 2/15 32/15
17 2 2/15 34/15
18 3 3/15 54/15
19 2 2/15 38/15
20 2 2/15 40/15
21 1 1/15 21/15
22 1 1/15 22/15

c. Solve for the mean of the sampling
distribution of the sample means by
using the following formula
= Σ[P()] ¿
270
15 ¿ 18
Therefore, the mean of the sampling
distribution of the sample mean is 18.

3.) The population mean is
18 and the mean of the
sample mean is 18.
Therefore,

4.) To solve for the variance () and the standard
deviation (σ) of the population, the given table
below will be utilized.
8
12
16
20
24
28
8 64
12 144
16 256
20 400
24 576
28 784

Variance: Standard Deviation:
𝜎2
=
∑ 𝑥2
𝑁
−(∑ 𝑥
𝑁 )
2
𝜎
2
=
2224
6
−(108
6 )
2
𝜎 2
= 4 6 .67
𝜎
2
=
√∑𝑥
2
𝑁
−(∑𝑥
𝑁 )
2
𝜎2
=
√2224
6
−(108
6 )
2
𝜎 2
=6.83

5) To solve for the variance () and the standard
deviation () of the sampling distribution of
the sample means, the following steps are
to be considered.
a. The following formula will be utilized
to solve for the variance () and the standard
deviation () of the sampling distribution of
the sample means.
𝜎
2
𝑥=√Σ[𝑥
2
• P(𝑥)]−𝝁
2
𝑥

b.) Construct the sampling distribution table for the sample
means, square the sample means and multiply the result to
the probabilities.
Sample Mean () Probability P() P()
14
15
16
17
18
19
20
21
22
14 196
15 225
16 256
17 289
18 324
19 361
20 400
21 441
22 484
14 196 1/15
15 225 1/15
16 256 2/15
17 289 2/15
18 324 3/15
19 361 2/15
20 400 2/15
21 441 1/15
22 484 1/15
14 196 1/15 196/15
15 225 1/15 225/15
16 256 2/15 512/15
17 289 2/15 578/15
18 324 3/15 972/15
19 361 2/15 722/15
20 400 2/15 800/15
21 441 1/15 441/15
22 484 1/15 484/15

c.) Solve for the variance and standard deviation of
the sampling distribution of the sample means by
using the given formula.
𝜎𝑥=√Σ[𝑥2
• P(𝑥)]−𝝁2
𝑥
𝜎 2
𝑥=3 28.67 −¿
𝜎 2
𝑥=4 .67
𝜎 𝑥=√328.67 −¿¿
𝜎 𝑥=2.16
Therefore, the variance and the standard deviation
of the sampling distribution of the sample means are
4.67 and 2.16, respectively.

6) The variance of the population and
sample mean are not equal with the
value of 46.67 and 4.67. The
standard deviation of the population
is 6.83 and standard deviation of the
sample means is 2.16. Therefore, σ ≠
.

Lesson 5.1.-_- Finding Variance and Standard Deviation of sampling distribution.pptx

Lesson 5.1.-_- Finding Variance and Standard Deviation of sampling distribution.pptx

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