Question

Assume that

.333.3​%

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

1488

​adults,

534

have sleepwalked.a. Assuming that the rate of

333.3​%

is​ correct, find the probability that

534

or more of the

1488

adults have sleepwalked.b. Is that result of

534

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

333.3​%?

a. Assuming that the rate of

333.3​%

is​ correct, the probability that

534

or more of the

1488

adults have sleepwalked is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

mean = n*p = 1488*0.3333 = 495.9504

SD = [n*p*(1-p)]^0.5 = [1488*0.3333*0.6667]^0.5 = 18.1838

a.

x = no. of people in 1488 people have sleepwalked

P(x >= 534) :

P(x>534) = 0.0182

b.

P(x>534) is too less so the result of 534 or more is significantly high

it suggests that the rate of 33.33% is less compared to the results in our sample.

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P(z<Z) table :

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