The amount of snow fall falling in a certian mountain range is normally distributed with a mean of 83 inches, and a standard deviation of 16 inches. What is the probability that the mean annual snowfall during 64 randomly picked years will exceed 85.8 inches?
Use this formula to find z:
z = (x - mean)/(sd/√n)
With your data:
z = (85.8 - 83)/(16/√64) = ?
Once you have the z-score, look at the z-table to determine your probability.
I hope this will help get you started.
.9192
To find the probability that the mean annual snowfall during 64 randomly picked years will exceed 85.8 inches, we need to use the central limit theorem and convert the problem into a standard normal distribution.
Step 1: Find the sample mean
The sample mean is equal to the population mean, which is 83 inches.
Step 2: Find the sample standard deviation
The sample standard deviation is equal to the population standard deviation divided by the square root of the sample size:
Standard deviation = 16 inches / √64 = 16 inches / 8 = 2 inches.
Step 3: Standardize the sample mean
To standardize the sample mean, we use the formula:
Z = (X - μ) / (σ / √n)
where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, X = 85.8 inches, μ = 83 inches, σ = 16 inches, and n = 64:
Z = (85.8 - 83) / (16 / √64) = 2.8 / 2 = 1.4.
Step 4: Find the probability using the standard normal distribution
Using a standard normal distribution table or a calculator, we can find the probability corresponding to a Z-score of 1.4. From the table, we find that the probability is approximately 0.9192.
Therefore, the probability that the mean annual snowfall during 64 randomly picked years will exceed 85.8 inches is approximately 0.9192 or 91.92%.
To find the probability that the mean annual snowfall during 64 randomly picked years will exceed 85.8 inches, we can use the Central Limit Theorem.
The Central Limit Theorem states that the distribution of sample means approaches a normal distribution, regardless of the shape of the original population, as the sample size increases.
In this case, we are given that the original distribution of snowfall is normally distributed with a mean of 83 inches and a standard deviation of 16 inches.
The mean of the sample means (also known as the sample mean) will still be 83 inches. However, the standard deviation of the sample means (also known as the standard error) will be the population standard deviation divided by the square root of the sample size.
The standard error = standard deviation / sqrt(sample size)
= 16 / sqrt(64)
= 16 / 8
= 2
Now, we need to calculate the z-score for the given value of 85.8 inches.
The z-score formula is:
z = (x - mean) / standard deviation
z = (85.8 - 83) / 2
= 2.8 / 2
= 1.4
Using a z-table or a statistical calculator, we can find that the probability of a z-score of 1.4 or higher is approximately 0.0808.
Therefore, the probability that the mean annual snowfall during 64 randomly picked years will exceed 85.8 inches is approximately 0.0808 or 8.08%.