Assume that 21.6% of people have sleepwalked. assume that in a random sample of 1487 adults, 359 have sleepwalked. a. assuming that the rate of 21.6% is correct, find the probability that 359 or more of the 1487 adults have sleepwalked. b. is that result of 359 or more significantly high? c. what does the result suggest about the rate of 21.6%?

a. The probability that 359 or more of 1487 adults sleepwalked is zero (0).

b. Yes, it's significantly higher than the 21.6% of people who sleepwalked.

c. The result suggests 321 adults.

Step-by-step explanation:

a. The assumed 359 sleepwalked adults exceed the 21.6% correct rate. 359 sleepwalked adults give 24.1%.

b. Yes, it is significantly high because it's 24.1% instead of correctly assumed 21.6%.

c. 21.6% sleepwalked adults give 321.2 people. Hence, people can only be in whole of numbers. Therefore, it is 321 adults.

the revenue of the fitness center, y, is the product of two factors: the number of members and the fee paid by each member. there are 320 members now, but number of members goes down by 10 for each $5 increase in fees, so if x is the number of $5 price increases then members is equal to (320 - 10x). since the fee is currently $45, and the fee increases by $5 for every price bump x, the fee is given by (45 + 5x). so, revenue y is given by y = (320 - 10x)(45 + 5x).