Answer to Question #231653 in Statistics and Probability for Jamilah

Question

Compute and plot a bar graph of the probability distribution of recoveries
out of 6 randomly selected patients.
[5 marks]
(iv) Plot the distribution function of the number of recoveries out of 6 randomly
selected patients.
[2 marks]
b) Phone calls enter the ”support desk” of an electricity supplying company on the
average two every 3 minutes. If one assumes an approximate Poisson process:
(i) What is the probability of no calls in 3 minutes?
[2 marks]
(ii) What is the probability of utmost 6 calls in a 9 minute period?
[2 marks]
c) Suppose the earnings of a laborer, denoted by X, are given by the following
probability distribution.
xi
P(xi)
0
8/27
1
12/27
2
6/27
3
1/27
Find the laborer’s expected earnings and the variance of his earnings.
[4 marks]

Answer 1

Part a

Arrange the data from the largest to smallest according to the frequency

Draw and label the x and y axes and then the bars corresponding to the frequencies. The Pareto chart shows that Florida has the highest number. The number is more than twice as high as the cost for Indiana.

Part iv

Part b

i)

$P(X≤2) = \sum_0^2 \frac{6^xe^{-6}}{x!}\\ = \frac{6^xe^{-6}}{0!}+ \frac{6^xe^{-6}}{1!}+ \frac{6^xe^{-6}}{2!}\\ =e^{-6}[\frac{6^0}{0!}+ \frac{6^1}{1!}+ \frac{6^2}{2!}]\\ =e^{-6}[1+ 6+ \frac{36}{2}]\\ =0.06195$

ii)

$P(6≤X≤9) = \sum_6^9 \frac{6^xe^{-6}}{x!}\\ = \frac{6^xe^{-6}}{6!}+ \frac{6^xe^{-6}}{7!}+ \frac{6^xe^{-6}}{8!}+ \frac{6^xe^{-6}}{9!}\\ =e^{-6}[\frac{6^6}{6!}+ \frac{6^7}{7!}+ \frac{6^8}{8!}+ \frac{6^9}{9!}]\\ =e^{-6}[64.8+ 55.5+ 41.7+27.8]\\ =0.470467$

Part c

Mean

$\mu= \frac{0*\frac{8}{27}+1* \frac{2}{27}+2* \frac{6}{27}+3*\frac{1}{27}}{\frac{8}{27}+ \frac{2}{27}+\frac{6}{27}+\frac{1}{27}}\\ \mu=\frac{\frac{17}{27}}{\frac{17}{27}}\\ \mu=1$

Variance

$\sigma= \frac{\frac{8}{27}*(0-1)^2+\frac{2}{27}*(1-1)^2+\frac{6}{27}*(2-1)^2+\frac{1}{27}*(3-1)^2}{\frac{8}{27}+ \frac{2}{27}+\frac{6}{27}+\frac{1}{27}}\\ \sigma =\frac{\frac{2}{3}}{\frac{17}{27}}\\ \sigma= \frac{18}{17}\\ \sigma=1.0588235$