Question

Assume that 33.4% of people have sleepwalked. Assume that in a random sample of 1459 ​adults, 526 have sleepwalked.

a. Assuming that the rate of 33.4% is​ correct, find the probability that 526 or more of the 1459 adults have sleepwalked.

b. Is that result of 526 or more significantly​ high?

c. What does the result suggest about the rate of 33.4?

Answer 1

To find the probability that 526 or more of the 1459 adults have sleepwalked, use the binomial probability formula and compare it to a significance level. The result suggests that the rate of 33.4% may be accurate.

Step-by-step explanation:

To find the probability that 526 or more of the 1459 adults have sleepwalked, we can use the binomial probability formula. The probability of success (p) is 0.334, the number of trials (n) is 1459, and the number of successes (x) is 526 or more. We can calculate the probability using a binomial calculator or statistical software.

To determine if the result of 526 or more is significantly high, we need to compare it to a significance level, typically 0.05. If the probability is less than or equal to the significance level, we can conclude that the result is significantly high.

The result suggests that the rate of 33.4% may be an accurate representation of the population, as 526 sleepwalkers out of 1459 falls within the expected range.

Answer 2

Step-by-step explanation:

Given that 33.4% of people have sleepwalked.

Sample size n =1459

Sample favourable persons = 526

Sample proportion p =
`(526)/(1459) \\=0.361`

Sample proportion p is normal for large samples with mean = 0.334 and

std error =
`\sqrt{(pq)/(n) } \\=\sqrt{(0.334(1-0.334))/(1459) } \\=0.0123`

a) P(526 or more of the 1459 adults have sleepwalked.)

`=P(p\geq 0.361)\\=P(Z\geq (0.361-0.334)/(0.0123) \\=P(Z\geq 2.20)\\=0.5-0.4861\\=0.0139`

b) Yes, because hardly 1.4% is the probability

c) 33.4 is very less compared to the average. Either sample should be improved representing the population or population mean should be increased accordingly.