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

Answer:

Step-by-step explanation:

Given that  33.4% of people have sleepwalked.

Sample size n =1459

Sample favourable persons = 526

Sample proportion p = $\frac{526}{1459} \\=0.361$

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

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

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

$=P(p\geq 0.361)\\=P(Z\geq \frac{0.361-0.334}{0.0123} \\=P(Z\geq 2.20)\\=0.5-0.4861\\=0.0139$

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.