We will now find the standard deviation. Recall that o = V npg is the standard deviation for the random variable r, where q = 1 – p is the probability of failure on a single trial. We previously determined that n = 10 and p = 0.2. So, q = Substitute the values for n, p and q into the formula and round the final answer to two decimal places. O = V npa V 10(0.2)( 0.8 0.8 реople
We will now find the standard deviation. Recall that o = V npg is the standard deviation for the random variable r, where q = 1 – p is the probability of failure on a single trial. We previously determined that n = 10 and p = 0.2. So, q = Substitute the values for n, p and q into the formula and round the final answer to two decimal places. O = V npa V 10(0.2)( 0.8 0.8 реople
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
screenshots are for two different problems
![Step 4
We will now find the standard deviation. Recall that o =
V npg is the standard deviation for the random
variable r, where q = 1 - p is the probability of failure on a single trial. We previously determined that n = 10
and p
= 0.2. So, q =
Substitute the values for n, p and q into the formula and round the final answer to two decimal places.
O =
npq
V 10(0.2)( 0.8
0.8
people
Submit
Skip (you cannot come back)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffddd0979-d40b-4561-8bba-cd60b353b254%2F861da222-8645-45ab-ae5e-bf90ca4b5e4a%2Fy6stvtb_processed.png&w=3840&q=75)
Transcribed Image Text:Step 4
We will now find the standard deviation. Recall that o =
V npg is the standard deviation for the random
variable r, where q = 1 - p is the probability of failure on a single trial. We previously determined that n = 10
and p
= 0.2. So, q =
Substitute the values for n, p and q into the formula and round the final answer to two decimal places.
O =
npq
V 10(0.2)( 0.8
0.8
people
Submit
Skip (you cannot come back)
![Step 2
Now, use the data in the table to make a histogram of the probability distribution.
1
2
3
4 or more
P(x)
0.46
0.33
0.17
0.03
0.01
Recall that the height of each bar in a histogram indicates the class frequency. So to construct the histogram,
put the x values on the horizontal axis and the probabilities on the vertical
discrete, each class mark will be centered over the x value. There are five classes in the table, so there are
axis. Since the data are
6.
X vertical bars on the histogram. Select the histogram of the probability distribution.
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
1
4 or more
1
2
4 or more
X
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
1
3 4 or more
1
2
3
4 or more
Submit
Skip (you cannot come back)
(x)d
(х)d
2.
(x)d
(х)d](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffddd0979-d40b-4561-8bba-cd60b353b254%2F861da222-8645-45ab-ae5e-bf90ca4b5e4a%2F6epxo4_processed.png&w=3840&q=75)
Transcribed Image Text:Step 2
Now, use the data in the table to make a histogram of the probability distribution.
1
2
3
4 or more
P(x)
0.46
0.33
0.17
0.03
0.01
Recall that the height of each bar in a histogram indicates the class frequency. So to construct the histogram,
put the x values on the horizontal axis and the probabilities on the vertical
discrete, each class mark will be centered over the x value. There are five classes in the table, so there are
axis. Since the data are
6.
X vertical bars on the histogram. Select the histogram of the probability distribution.
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
1
4 or more
1
2
4 or more
X
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
1
3 4 or more
1
2
3
4 or more
Submit
Skip (you cannot come back)
(x)d
(х)d
2.
(x)d
(х)d
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