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Answer :
The answer is e. none of the above
To solve this problem, we'll use the properties of the normal distribution and the central limit theorem.
Given:
- Mean (μ) = 98 inches
- Standard Deviation (σ) = 16 inches
- Sample size (n) = 64
- Desired snowfall threshold (x) = 100.8 inches
First, we'll find the standard error of the mean (SEM), which is the standard deviation of the sampling distribution of the sample mean:
SEM = σ / √n
SEM = 16 / √64 = 16 / 8 = 2
Next, we'll find the z-score corresponding to the desired snowfall threshold:
z = (x - μ) / SEM
z = (100.8 - 98) / 2 = 2.8 / 2 = 1.4
Now, we'll use a standard normal distribution table or calculator to find the probability that a z-score is greater than 1.4. This represents the probability that the mean annual snowfall during 64 randomly picked years will exceed 100.8 inches.
From the standard normal distribution table, the probability corresponding to a z-score of 1.4 is approximately 0.9192.
However, we want the probability that the z-score is greater than 1.4. Since the normal distribution is symmetric, we can subtract the cumulative probability from 1:
P(Z > 1.4) = 1 - P(Z ≤ 1.4)
Now, looking up 1.4 in the standard normal distribution table, we find the cumulative probability to be approximately 0.9192.
So, P(Z > 1.4) = 1 - 0.9192 = 0.0808
Therefore, the probability that the mean annual snowfall during 64 randomly picked years will exceed 100.8 inches is approximately 0.0808.
But this is just the probability for one-tail (the tail above the mean). However, if you're given multiple-choice answers, they might be for two-tail probabilities. To convert this to a two-tail probability, we double the value:
P(Z > 1.4) [tex]*[/tex] 2 = 0.0808 [tex]*[/tex] 2 = 0.1616
so the answer is e. none of the above
Complete question is here :
The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 98 inches and a standard deviation of 16 inches. What is the probability that the mean annual snowfall during 64 randomly picked years will exceed 100.8 inches?
a. 0.1591
b. 0.2400
c. 0.3173
d. 0.4404
e. None of the above
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