The percentage for bacterial infection fatalities in local hospitals for each U.

Question

The percentage for bacterial infection fatalities in local hospitals for each U.S. state was approximately normally distributed with a mean=56.9%andstdev=6.8%(total fatalities-26173 in 1997.) In 2011, there were 10,076 fatalitie which is less than 40% of 1997 total with a mean fatalty rate due to bacterial infection in all states=30.8%andstdev=5.1%. There is an assumption of normal distribution of fatalities in all states.

1. What number fo states in 2011 would you expect to have more than 50% of their traffice fatalities from bacterial infections? Draw a picture to supplement your answer.

2. How does your answer to 1. compare to the number of states in 1997 expected to have more than 50% of fatalities due to bacterial infection? (estimate this for 1997 and compare. Decide whether you need to use the 5 step hypothesis testing or not.

Explanation / Answer

p be the percentage of fatalties in bacterial infection.

In 2011, p follows normal with mean = 30.8% = 0.308 and standard deviation = 5.1% = 0.051

So Probability that percentage fatalties is greater than 50% = P(p> 0.5)

= P( (p-0.308)/0.051 > (0.5-0.308)/0.051)

= P(Z > 0.192/0.051)

= 1-P(Z<=3.7647) where Z follows N(0,1)

= 1- NORMDIST(3.7647,0,1,TRUE)

= 0.00008

So the number of states in 2011 where one should expect more than 50% fatalties = number of states * above probability

= 50*0.00008= 0.0042 = 0 (approximately)

Similarly the required probability in 1997

P(p> 0.5) , note that in 1997 p follows normal with mean = 56.9% =0.569 and standard deviation = 0.068

= P( (p-0.569)/0.068 > (0.5-0.569)/0.068)

= P(Z > -0.069/0.068)

= 1-P(Z<= -1.0147) where Z follows N(0,1)

=P(Z<=1.0147)

= NORMDIST(1.0147,0,1,TRUE)

= 0.1551

So the number of states in 1997 where one should expect more than 50% fatalties = number of states * above probability

= 50*0.1551= 7.76 = 8 (approximately)

So in 2011 the number of states in which number of fatalties due to bacterial infection are more than 50% is reduced to 0 from 8. Which is a quite reasonable achievement to control bacterial infection

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