Solution Found!
The information below is the number of daily emergency
Chapter 6, Problem 5E(choose chapter or problem)
The information below is the number of daily emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were 2 emergency calls, and 9 days on which there were 3 emergency calls.
\(\begin{array}{|cc|}
\hline \text { Number of Calls } & \text { Frequency } \\
\hline 0 & 8 \\
1 & 10 \\
2 & 22 \\
3 & 9 \\
4 & 1 \\
\text { Total } & 50 \\
\hline
\end{array}\)
a. Convert this information on the number of calls to a probability distribution.
b. Is this an example of a discrete or continuous probability distribution?
c. What is the mean number of emergency calls per day?
d. What is the standard deviation of the number of calls made daily?
Questions & Answers
QUESTION:
The information below is the number of daily emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were 2 emergency calls, and 9 days on which there were 3 emergency calls.
\(\begin{array}{|cc|}
\hline \text { Number of Calls } & \text { Frequency } \\
\hline 0 & 8 \\
1 & 10 \\
2 & 22 \\
3 & 9 \\
4 & 1 \\
\text { Total } & 50 \\
\hline
\end{array}\)
a. Convert this information on the number of calls to a probability distribution.
b. Is this an example of a discrete or continuous probability distribution?
c. What is the mean number of emergency calls per day?
d. What is the standard deviation of the number of calls made daily?
ANSWER:Step 1 of 4
Given the number of daily emergency service calls.
Then the table is given below.
|
Number of Calls |
Frequency |
|
0 |
8 |
|
1 |
10 |
|
2 |
22 |
|
3 |
9 |
|
4 |
1 |
|
Total |
50 |
Our goal is:
a). We need to convert the number of calls to a probability distribution.
b). We need to find this example is discrete or continuous probability distribution.
c). We need to find the mean.
d). We need to find the standard deviation.
a). Now we have to convert the number of calls to a probability distribution.
Then the table is given below.
|
Number of Calls (X) |
Frequency |
P(X) |
\(\mu=\mathrm{X} \times \mathrm{P}(\mathrm{X})\) |
\(\sigma^{2}=(X-\mu)^{2} \times P(X)\) |
|
0 |
8 |
\(\frac{8}{50}=0.16\) |
\(0 \times 0.16=0\) |
0.4624 |
|
1 |
10 |
\(\frac{10}{50}=0.2\) |
\(1 \times 0.2=0.2\) |
0.098 |
|
2 |
22 |
\(\frac{22}{50}=0.44\) |
\(2 \times 0.44=0.88\) |
0.0396 |
|
3 |
9 |
\(\frac{9}{50}=0.18\) |
\(3 \times 0.18=0.54\) |
0.3042 |
|
4 |
1 |
\(\frac{1}{50}=0.02\) |
\(4 \times 0.02=0.08\) |
0.1058 |
|
Total |
50 |
|
\(\mu=1.7\) |
\(\sigma^{2}=1.01\) |
From the above table mean is 1.7 and the variance is 1.01.