find the indicated prob.:

the amount of snowfall falling in a certain mountain range is normally distributed with a mean of 86 inches, and a standard deviation of 12 inches. what is the probability that the mean annual snowfall during 36 randomly picked years will exceeds 88.8 inches?

0.0808

how did u come up with it?thanks

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.

Write the z score rounded to 2 places and the probability as a decimal rounded to 4 places

To find the probability that the mean annual snowfall during 36 randomly picked years will exceed 88.8 inches, we can use the Central Limit Theorem and the properties of the normal distribution.

The Central Limit Theorem states that, for a large enough sample size, the distribution of sample means will approach a normal distribution, regardless of the shape of the original population.

To apply the Central Limit Theorem, we need to calculate the mean and standard deviation of the sampling distribution of sample means.

The mean of the sampling distribution of sample means is equal to the mean of the population, which is given as 86 inches.

The standard deviation of the sampling distribution of sample means, also known as the standard error, can be found using the formula: standard deviation / square root of sample size.

In this case, the standard deviation is 12 inches, and the sample size is 36 years. Therefore, the standard error is 12 / √36 = 12 / 6 = 2 inches.

Now, we have the mean (86 inches) and the standard error (2 inches) of the sampling distribution. To find the probability that the mean annual snowfall during 36 randomly picked years will exceed 88.8 inches, we need to convert this value into a z-score.

The formula to calculate the z-score is: (x - μ) / σ, where x is the value we want to convert, μ is the mean, and σ is the standard error.

In this case, x = 88.8 inches, μ = 86 inches, and σ = 2 inches. Plugging these values into the formula, we get: (88.8 - 86) / 2 = 2.8 / 2 = 1.4.

Next, we need to use a z-table or a calculator to find the probability associated with a z-score of 1.4. Assuming a standard normal distribution (mean = 0, standard deviation = 1), the probability can be determined from the z-table as approximately 0.9192.

Therefore, the probability that the mean annual snowfall during 36 randomly picked years will exceed 88.8 inches is approximately 0.9192, or 91.92%.