In: Math
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. 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 5.9% 9.1% 19.1% 29.9% 25.5% 8.8% 1.7% (a) Using the age midpoints x and the percent of nurses, do we have a valid probability distribution? Explain. No. The events are indistinct and the probabilities do not sum to 1. Yes. The events are distinct and the probabilities sum to 1. No. The events are indistinct and the probabilities sum to 1. Yes. The events are distinct and the probabilities do not sum to 1. (b) Use a histogram to graph the probability distribution in part (a). Maple Generated Plot Maple Generated Plot Maple Generated Plot Maple Generated Plot (c) Find the probability that a British nurse selected at random in 1851 would be 60 years of age or older. (Round your answer to three decimal places.) (d) Compute the expected age μ of a British nurse contemporary to Florence Nightingale. (Round your answer to two decimal places.) yr (e) Compute the standard deviation σ for ages of nurses shown in the distribution. (Round your answer to two decimal places.) yr
Part a.
Yes. The events are distinct and the probabilities sum to 1.
Part b.
The histogram is obtained as follows:
Instructions to build the histogram is given below:
Part C.
P(60 years older) =P(60-69)+(P(70-79)+P(80+)
=0.25+0.088+0.017
=0.36
Thus, the probability that British nurse age would be 60 years or older is 0.36.
Part d.
Midpoint x | Percent of nurses P(x) | x-P(X) |
24.5 | 5.90% | 1.45 |
34.5 | 9.10% | 3.14 |
44.5 | 19.10% | 8.50 |
54.5 | 29.90% | 16.30 |
64.5 | 25.50% | 16.45 |
74.5 | 8.80% | 6.56 |
84.5 | 1.70% | 1.44 |
∑x-P(X) | 53.82 |
Thus, the expected age of British nurse come temporary to Florence nightingale is 53.82.
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