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 determine the likelihood of the mean annual precipitation during 25 randomly picked years being less than 90.8 inches, we need to use the properties of the normal distribution.
Step 1: Understand the problem
We are given that the annual precipitation amounts in the mountain range are normally distributed with a mean of 88 inches and a standard deviation of 10 inches. We want to find the likelihood of the mean annual precipitation during 25 years being less than 90.8 inches.
Step 2: Convert the problem into a standard normal distribution
To do this, we will use the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size.
Standard error of the mean (SE) = standard deviation / √(sample size)
SE = 10 / √(25) = 10 / 5 = 2
Now we have a standard normal distribution with a mean of 0 (since we converted to a standard normal distribution) and a standard deviation of 1 (since we divided the standard deviation by itself, which resulted in 1).
Step 3: Calculate the z-score
The z-score is a measure of how many standard deviations an observed value is from the mean. In this case, the observed value is 90.8 inches.
z = (observed value - mean) / standard error
z = (90.8 - 88) / 2 = 2.8 / 2 = 1.4
Step 4: Find the probability using the z-table
The z-table provides the probability of obtaining a z-score. We need to find the probability of obtaining a z-score less than 1.4.
Looking up the z-score of 1.4 in the z-table, we find that the probability is approximately 0.9192.
Therefore, the likelihood of the mean annual precipitation during 25 randomly picked years being less than 90.8 inches is approximately 0.9192, or 91.92%.