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?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
78
To find the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 90.8 inches, we need to calculate the probability using the normal distribution.
1. Start by converting the problem into the standard normal distribution, which has a mean of 0 and a standard deviation of 1. We can do this using the formula for the standard normal distribution:
z = (x - μ) / σ
where z is the z-score, x is the observed value, μ is the mean, and σ is the standard deviation.
2. Calculate the z-score for the observed value of 90.8 inches using the given mean and standard deviation:
z = (90.8 - 88) / 10
3. Look up the probability associated with the z-score in the standard normal distribution table. The table provides the area under the curve to the left of a given z-score. In this case, we want to find the area to the left of the calculated z-score.
4. The probability associated with the z-score can be read directly from the table or calculated using statistical software. For example, using a standard normal distribution table or a calculator, the probability of obtaining a z-score less than the calculated value of 0.28 is approximately 0.6103.
5. However, we are interested in the probability of obtaining a mean annual precipitation less than 90.8 inches over 25 years. Since we are dealing with the mean of 25 randomly picked years, we need to consider the sampling distribution of the mean.
6. The Central Limit Theorem states that if a sample size is large enough (typically n > 30), the sampling distribution of the mean will be approximately normal, regardless of the shape of the population distribution. In this case, we can assume that the sampling distribution of the mean will be approximately normal.
7. For the sampling distribution of the mean, the mean of the sampling distribution is equal to the population mean, and the standard deviation is equal to the population standard deviation divided by the square root of the sample size, or σ / sqrt(n).
8. Substitute the values into the formula:
mean of the sampling distribution = μ = 88 inches
standard deviation of the sampling distribution = σ / sqrt(n) = 10 inches / sqrt(25) = 2 inches
9. Repeat steps 1-4 with the new values for the sampling distribution (mean and standard deviation) to find the probability of obtaining a mean annual precipitation less than 90.8 inches over 25 years.
The cumulative probability obtained from step 9 will give you the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 90.8 inches.