Please someone help i’ll give brainliest

Suppose that the annual rainfall in Ferndale, California, is known to be normally distributed, with a mean of 35.5 inches and a

erastovalidia [21]

The average annual rainfall will be more than 40.6 inches in about 2.1 percent of the years.

<h3>What is the average annual rainfall?</h3>

Generally, the equation for the probability is mathematically given as

$P( X < x) = p( Z < x - \mu / \sigma)$

Therefore

$P( X > x) = 0.021$

$P( X < x ) = 1 - 0.021\\\\\p( X < x) = 0.979$

$P( Z < x - \mu / \sigma) = 0.979$

The z score for the probability of 0.979 is 2.034, according to the table of z values.

$x - \mu / \sigma \\\\x= 2.034$

In the given equation, replace the values of mu and sigma with their respective values, and then solve for x.

$x - 35.5 / 2.5 \\x= 2.034$

In conclusion, The average annual rainfall will be more than 40.6 inches in about 2.1 percent of the years.

Read more about average

brainly.com/question/24057012

#SPJ1

7 0

2 years ago

Data from the Centers for Disease Control and Prevention indicate that weights of American adults in 2005 had a mean of 167 poun

sesenic [268]

Answer:

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .