What is equivalent to 8/9 the denominator is 72

Please helo i will make brainless for correct answers!

elena-s [515]

False is the correct answer

7 0

3 years ago

Assume that 24.5% of people have sleepwalked. Assume that in a random sample of 1478 adults, 369 have sleepwalked. a. Assuming t

solniwko [45]

Answer:

a) 0.3483 = 34.83% probability that 369 or more of the 1478 adults have sleepwalked.

b) 369 < 403.4, which means that 369 is less than 2.5 standard deviations above the mean, and thus, a result of 369 is not significantly high.

c) Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A result is considered significantly high if it is more than 2.5 standard deviations above the mean.

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .