Question

5.7 Aapproximately 6.8% (almost 2.4 million Canadians) were living with diabetes in 2008/09 [3). Assume that the situation has not changed since then. If there are 100 students in a biostatistics class, a) What is the chance that no more than one student is diabetic? b) What is the chance there is one diabetic student? c) What is the chance there is more than one diabetic student? d) What is the chance there are seven diabetic students? e) What is the chance there are seven or more diabetic students? f) What would be considered a "normal range" for the number of diabetic students in this class?

It is known that 1/3 of the St. Catharines adult population does not accumulate enough physical activity to experience health benefits. Lack of physical activity is a risk factor for various health conditions including cardiovascular disease, cerebrovascular disease, and osteoporosis. If 10 adults are randomly selected from this population:

1. a) What is the probability that exactly 8 of them will not be physically active?

2. b) What is the probability that at least 9 will not be physically active?

3. c) Find mean and standard deviation for the number of adults not being physically active

4. d) Would it be unusual to survey 10adults and find that 5 of them are not physically active? Why or why not?

Answer 1

5.7) X= no of students out of 100, who are diabetic

X has a binomial distribution with parameters 100 and .068

a) p(x<=1)=.00725

b) p(x=1)=.00638

c) p(x>1)=1-.00725=.99275

d)p(x=7)=.15402

e) p(x>7)= .524403

f) normal range is (mean-3sd,mean+3sd)

Here mean= 100*.068=6.8

Sd=square root of 100*.068*(1-.068), which is 2.517459

Hence the normal range is (0,14.35238)

The value of mean-3sd is negative bit the number of students is atlrast non negative and hence we replace the lower limit by 0.

Second problem

X= no. of individuals in the sample of 10 who are not physically acive

Then X~ binomial(10,1/3)

A. P(x=8)=.00305

B.P(x>=9)=p(x=9)+p(x=10)=.000356

C. Mean=10*1/3=3.33333

Sd=square root of 10*(1/3)*(2/3), which is 1.490712

D.p(x=5)=.13656. Thus the chance of observing 5 out of 10 is .13656, which is moderately higher. Hence the event is not unusual.

For query in above, comment.