Question

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

Answer 1

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