In: Statistics and Probability
Age range (yr) | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80+ |
Midpoint x | 24.5 | 34.5 | 44.5 | 54.5 | 64.5 | 74.5 | 84.5 |
Percent of nurses | 5.4% | 9.8% | 19.8% | 29.2% | 25.5% | 8.7% | 1.6% |
What was the age distribution of nurses in Great Britain at the time of Florence Nightingale? Suppose we have the following information. Note: In 1851 there were 25,466 nurses in Great Britain.
(e) Compute the standard deviation σ for ages of nurses shown in the distribution. (Round your answer to two decimal places.)
Solution :
x | P(x) | x * P(x) | x2 * P(x) |
24.5 | 0.054 | 1.323 | 32.4135 |
34.5 | 0.098 | 3.381 | 116.6445 |
44.5 | 0.198 | 8.811 | 392.0895 |
54.5 | 0.292 | 15.914 | 867.313 |
64.5 | 0.255 | 16.4475 | 1060.864 |
74.5 | 0.087 | 6.4815 | 482.8718 |
84.5 | 0.016 | 1.352 | 114.244 |
Sum | 1 | 53.71 | 3066.44 |
Mean = = X * P(X) = 53.71
Standard deviation =
=X 2 * P(X) - 2
= 3066.44 - 53.712
= 13.48