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statprob3

Page history last edited by PBworks 18 years ago

Problem 3 of 3

 

It is found that 40% of all students returning to school in September catch a common cold, while the rest do not. From among the students in the school 600 are randomly selected.

 

(a) Determine the mean and population standard deviation for the students who catch a cold.

 

(b) What is the 90% confidence interval for the mean number of students who will catch a cold in this group. Explain the meaning of your confidence interval.

 

Solution

 

Answer to Part A

 

First you list the key information:

- 40% catch a cold

- 600 random students selected

 

So this means that:

n = 600 * number of trails

p = 0.4 * probability of 'success'

q = 0.6 * probability of 'failure'

 

From the above, you can now find the mean and the standard deviation:

 

To find the Mean:

μ = np = 600(0.4) = 240 students

 

To find the Standard Deviation:

σ = √npq

= √(600)(0.4)(0.6)

= 12 students

 

The diagram below shows the mean and standard deviation with the '68-95-99 Rule':

 

Answer to Part B

 

Note; To find the value that I had to use to calculate the interval, I looked at the chart for finding: "Areas under a standard normal curve to the left of z". UTILITY 29 in textbook.

 

90% = μ ± 1.70σ

= 240 ± 1.70(12)

= 240 ± 20.4

 

Minimum value: 240 - 20.4 = 219.6

Maximum value: 240 + 20.4 = 260.4

 

So the interval would be written as: [ 219.6 , 260.4 ]

But the question is talking about students and you can't have 'fractions of a person', and so you would round off to: [ 220.0 , 260.0 ]

 

This confidence interval shows that the calculation made is 90% sure 'confident' that the minimum number of students who are sick is 260 and the maximum number of students who are sick is 220 students.

 

 

 

 

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