Question

ssume that

32.632.6​%

of people have sleepwalked. Assume that in a random sample of

15011501

​adults,

532532

have sleepwalked.a. Assuming that the rate of

32.632.6​%

is​ correct, find the probability that

532532

or more of the

15011501

adults have sleepwalked.b. Is that result of

532532

or more significantly​ high?c. What does the result suggest about the rate of

32.632.6​%?

a. Assuming that the rate of

32.632.6​%

is​ correct, the probability that

532532

or more of the

15011501

adults have sleepwalked is

nothing .

Answer 1

Assume that 32.6​% of people have sleepwalked. Assume that in a random sample of 1501 adults,532 have sleepwalked.

a. Assuming that the rate of 32.6​% is​ correct, find the probability that 532 or more of the 1501 adults have sleepwalked.

b. Is that result of 532 or more significantly​ high?

c. What does the result suggest about the rate of 32.6​%?

μ = np = 1501*0.326 = 489.326

σ= √(npq) = √(1501*0.326*0.674)

=18.16055

a)

P(x>= 532) = 1- P(x<532)

=1-P(x<532-0.5) .....(using continuity correction)

=1-P(x<531.5)

For x=531.5

z=(531.5-489.326)/18.16055

=2.32

P(x<531.5) =P(z<2.32)

=0.9898

P(x>= 532) =1-P(x<531.5)

= 1-0.9898

=0.0102

b)

Since we got probability in part(a) as 0.0102 which is less than 0.05

So ,it is unusual event

Hence, Yes ,the result of 532 or more significantly​ high

c)

As we see that the probability result is significantly high,so rate of 32.6% is not valid

Thus,there is sufficient evidence to reject that claim that the rate is 32.6%