6) The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 88 inches, and a standard deviation of 10 inches. What is the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 90.8 inches?

To solve this problem, we can use the Central Limit Theorem and the properties of the normal distribution.

1) Calculate the standard error of the mean (SE) using the formula:
SE = σ / √n
where σ is the standard deviation (10 inches) and n is the sample size (25 years).

SE = 10 / √25
SE = 10 / 5
SE = 2 inches

2) Convert the value we want to find the likelihood for (90.8 inches) into a z-score using the formula:
z = (x - μ) / SE
where x is the value (90.8 inches), μ is the mean (88 inches), and SE is the standard error.

z = (90.8 - 88) / 2
z = 2.8 / 2
z = 1.4

3) Use a standard normal distribution table to find the area under the curve to the left of the z-score of 1.4. This represents the likelihood that the mean annual precipitation will be less than 90.8 inches.

Using the table, we find that the area to the left of 1.4 is approximately 0.9192.

4) Convert the decimal probability to a percentage by multiplying by 100:

0.9192 * 100 = 91.92%

Therefore, the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 90.8 inches is approximately 91.92%.

To find the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 90.8 inches, we can use the Central Limit Theorem.

The Central Limit Theorem states that the distribution of sample means will approach a normal distribution as the sample size increases, regardless of the shape of the population distribution.

In this problem, since the sample size is 25, we can assume that the distribution of sample means is approximately normal.

To calculate the z-score, which measures the number of standard deviations the mean is away from the population mean, we can use the formula:

z = (x - μ) / (σ / √n)

Where:
x = the value we want to find the likelihood for (90.8 inches),
μ = the population mean (88 inches),
σ = the population standard deviation (10 inches),
n = the sample size (25).

Plugging in the given values into the formula:

z = (90.8 - 88) / (10 / √25)
= 2.8 / 2
= 1.4

Now we need to find the probability of the mean annual precipitation being less than z = 1.4.

To find this probability, we can use a standard normal distribution table or a statistical calculator. Using a standard normal distribution table, we can look up the probability associated with z = 1.4, which is approximately 0.9192.

Therefore, the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 90.8 inches is approximately 0.9192, or 91.92%.

z = (x -mean)/(sd/sqrt(n))

z = (90.8-88)/(10/sqrt(25))

1.4