Question

# What was the age distribution of nurses in Great Britain at the time of Florence Nightingale?...

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 5.4% 9.5% 19.9% 29.2% 24.8% 9.7% 1.5% (a) Using the age midpoints x and the percent of nurses, do we have a valid probability distribution? Explain. Yes. The events are distinct 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. No. The events are indistinct 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

Yes. The events are distinct and the probabilities sum to 1.

b)

c)

probability that a British nurse selected at random in 1851 would be 60 years of age or older

=0.2480+0.097+0.0150 =0.3600

d)

 x P(X=x) xP(x) x2P(x) 24.5 0.0540 1.32300 32.41350 34.5 0.0950 3.27750 113.07375 44.5 0.1990 8.85550 394.06975 54.5 0.2920 15.91400 867.31300 64.5 0.2480 15.99600 1031.74200 74.5 0.0970 7.22650 538.37425 84.5 0.0150 1.26750 107.10375 total 53.8600 3084.0900 E(x) =μ= ΣxP(x) = 53.8600 E(x2) = Σx2P(x) = 3084.0900 Var(x)=σ2 = E(x2)-(E(x))2= 183.190400 std deviation= σ= √σ2 = 13.53478

expected age μ =53.86

e)

standard deviation σ =13.53

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