Question 1155142: The number of calls received per day at a crisis helpline is distributed according to the following discrete probability distribution. Find the missing number and then use your answer to compute the mean and standard deviation of the discrete probability distribution. Here, the discrete random variable, x, represents the number of calls in a day.
x 30 31 32 33 34
P(X) .05 .21 ? .15 .11
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Let y be the missing number for now.
The values in the probability P(X) row must add to 1
0.05+0.21+y+0.15+0.11 = 1
y+0.52 = 1
y = 1-0.52
y = 0.48
The missing number in the table is 0.48
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Computing the mean (mu)
I'll write the table out like this
| X | P(X) |
| 30 | 0.05 |
| 31 | 0.21 |
| 32 | 0.48 |
| 33 | 0.15 |
| 34 | 0.11 |
Now add on a X*P(X) column. Each row in this column is the result of multiplying the X and P(X) values for the given row. Eg: 31*0.21 = 6.51 for row 2
| X | P(X) | X*P(X) |
| 30 | 0.05 | 1.5 |
| 31 | 0.21 | 6.51 |
| 32 | 0.48 | 15.36 |
| 33 | 0.15 | 4.95 |
| 34 | 0.11 | 3.74 |
Adding up everything in the X*P(X) column yields
1.5+6.51+15.36+4.95+3.74 = 32.06
The mean is mu = 32.06
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Computing the standard deviation (sigma)
Subtract the mean mu = 32.06 from each data value X. This forms the (X-mu) column.
Then square each value in that column to get another column (X-mu)^2
Finally, multiply that with the P(X) column
This is what the table should look like
| X | P(X) | X-mu | (X-mu)^2 | (X-mu)^2*P(X) |
| 30 | 0.05 | -2.06 | 4.2436 | 0.21218 |
| 31 | 0.21 | -1.06 | 1.1236 | 0.235956 |
| 32 | 0.48 | -0.06 | 0.0036 | 0.001728 |
| 33 | 0.15 | 0.94 | 0.8836 | 0.13254 |
| 34 | 0.11 | 1.94 | 3.7636 | 0.413996 |
Add up the values in the last column
0.21218+0.235956+0.001728+0.13254+0.413996 = 0.9964
The variance is 0.9964
Take the square root of this to get sqrt(0.9964) = 0.9981984 approximately
sigma = 0.9981984 is the approximate standard deviation
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