The random variable X, representing the number of nuts in a chocolate bar has the following probability distribution. Compute the mean, variance and standard deviation and interpret the results. 2 4 P(X) 1/10 2/10 3/10 3/10 1/10 Solution: To find for the standard deviation, square the variance. That is: Following the steps in finding the variance and standard deviation of a discrete probability distribution, the following table will be made. (X – µ)². P(x) o = EI(X – H)² . P(X)] =V1.29 Px) X . P{x) (X – µ)² 1/10 -2.10 4.41 0.441 1 2/10 2/10 -1.10 o = 1.14 1.21 0.242 This means that for every chocolate bar, the number of nuts is in average of 2.10. The 2 3/10 6/10 -0.10 0.01 0.003 3 3/10 9/10 -0.90 0.81 0.243 variance is 1.29 and the standard deviation is 1.14. This means that the data points are very spread out from the mean, and from each other. 1/10 4/10 -1.90 3.61 0.361 H- 2.10 g2 = 1.29 Activity: Solve the following problems. 1. The number of cars sold per day at a local car dealership, along with its corresponding probabilities, is shown in the succeeding table. Compute the mean, variance and the standard deviation of the probability distribution by following the given steps. Interpret the results. 3 4 2/10 _X 2 P(X) 1/10 2/10 3/10 2/10 2. The number of items sold per day at a retail store, with its corresponding probabilities, is shown in the table. Find the mean, variance and standard deviation of the probability distribution. Interpret the results. 19 20 21 22 23 P(X) 0.20 0.20 0.30 0.20 0.10
The random variable X, representing the number of nuts in a chocolate bar has the following probability distribution. Compute the mean, variance and standard deviation and interpret the results. 2 4 P(X) 1/10 2/10 3/10 3/10 1/10 Solution: To find for the standard deviation, square the variance. That is: Following the steps in finding the variance and standard deviation of a discrete probability distribution, the following table will be made. (X – µ)². P(x) o = EI(X – H)² . P(X)] =V1.29 Px) X . P{x) (X – µ)² 1/10 -2.10 4.41 0.441 1 2/10 2/10 -1.10 o = 1.14 1.21 0.242 This means that for every chocolate bar, the number of nuts is in average of 2.10. The 2 3/10 6/10 -0.10 0.01 0.003 3 3/10 9/10 -0.90 0.81 0.243 variance is 1.29 and the standard deviation is 1.14. This means that the data points are very spread out from the mean, and from each other. 1/10 4/10 -1.90 3.61 0.361 H- 2.10 g2 = 1.29 Activity: Solve the following problems. 1. The number of cars sold per day at a local car dealership, along with its corresponding probabilities, is shown in the succeeding table. Compute the mean, variance and the standard deviation of the probability distribution by following the given steps. Interpret the results. 3 4 2/10 _X 2 P(X) 1/10 2/10 3/10 2/10 2. The number of items sold per day at a retail store, with its corresponding probabilities, is shown in the table. Find the mean, variance and standard deviation of the probability distribution. Interpret the results. 19 20 21 22 23 P(X) 0.20 0.20 0.30 0.20 0.10
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Question
![Example:
The random variable X, representing the number of nuts in a chocolate bar has the following probability
distribution. Compute the mean, variance and standard deviation and interpret the results.
4
1/10
P(X)
1/10
2/10
3/10
3/10
Solution:
To find for the standard deviation,
square the variance. That is:
Following the steps in finding the variance and standard deviation
of a discrete probability distribution, the following table will be made.
(X – 4)². Pıx)
o = EI(X – 1)² . P(X)]
=V1.29
o = 1.14
This means that for every chocolate bar,
the number of nuts is in average of 2.10. The
Px)
X . P(x)
(X – 4)?
1/10
-2.10
4.41
0.441
1
2/10
2/10
-1.10
1.21
0.242
2
3/10
6/10
-0.10
0.01
0.003
3/10
9/10
-0.90
0.81
0.243
variance is 1.29 and the standard deviation is
4
1/10
4/10
-1.90
3.61
0.361
1.14. This means that the data points are very
spread out from the mean, and from each
other.
H= 2.10
g2 = 1.29
Activity:
Solve the following problems.
1. The number of cars sold per day at a local car dealership, along with its corresponding probabilities, is shown in
the succeeding table. Compute the mean, variance and the standard deviation of the probability distribution by
following the given steps. Interpret the results.
_X
P(X)
2
3/10
3
2/10
4
1/10
2/10
2/10
2. The number of items sold per day at a retail store, with its corresponding probabilities, is shown in the table. Find
the mean, variance and standard deviation of the probability distribution. Interpret the results.
19
20
21
22
23
P(X)
0.20
0.20
0.30
0.20
0.10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0eec2776-ebf8-46e7-8e00-300b51bb2dd3%2Fd78d5000-045c-4c98-935e-9a19db016448%2Fdg8cy3r_processed.png&w=3840&q=75)
Transcribed Image Text:Example:
The random variable X, representing the number of nuts in a chocolate bar has the following probability
distribution. Compute the mean, variance and standard deviation and interpret the results.
4
1/10
P(X)
1/10
2/10
3/10
3/10
Solution:
To find for the standard deviation,
square the variance. That is:
Following the steps in finding the variance and standard deviation
of a discrete probability distribution, the following table will be made.
(X – 4)². Pıx)
o = EI(X – 1)² . P(X)]
=V1.29
o = 1.14
This means that for every chocolate bar,
the number of nuts is in average of 2.10. The
Px)
X . P(x)
(X – 4)?
1/10
-2.10
4.41
0.441
1
2/10
2/10
-1.10
1.21
0.242
2
3/10
6/10
-0.10
0.01
0.003
3/10
9/10
-0.90
0.81
0.243
variance is 1.29 and the standard deviation is
4
1/10
4/10
-1.90
3.61
0.361
1.14. This means that the data points are very
spread out from the mean, and from each
other.
H= 2.10
g2 = 1.29
Activity:
Solve the following problems.
1. The number of cars sold per day at a local car dealership, along with its corresponding probabilities, is shown in
the succeeding table. Compute the mean, variance and the standard deviation of the probability distribution by
following the given steps. Interpret the results.
_X
P(X)
2
3/10
3
2/10
4
1/10
2/10
2/10
2. The number of items sold per day at a retail store, with its corresponding probabilities, is shown in the table. Find
the mean, variance and standard deviation of the probability distribution. Interpret the results.
19
20
21
22
23
P(X)
0.20
0.20
0.30
0.20
0.10
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