The annual precipitation for one city is normally distributed with a mean of 72 inches and a standard deviation of 3.5 inches. In 95% of the years, the precipitation in this city is between___ and ___
From a normal distribution table, 97.5% probability corresponds to +1.96σ.
Thus μ±1.96σ corresponds to 95% of all years.
1.96σ = 1.96*3.5=6.86"
Therefore 95% of the years have precipitation between 72±6.86"
n1-25, n2-36, s1 power 2=50, s2 2 power=72, m1=100. m2=105
n1-25, n2-36, s1 power 2=50, s2 2 power=72, m1=100. m2=105
To find the range within which the precipitation lies in 95% of the years, we need to calculate the z-scores associated with the 95th percentile. Here's how you can do it:
1. Determine the z-score associated with the 95th percentile using a standard normal distribution table or a statistical calculator. For a two-tailed test, the z-score is approximately 1.96.
2. Now, we'll use the formula for converting a z-score back to the original scale of measurement:
x = μ + (z * σ)
where:
x = the value on the original scale
μ = the mean of the distribution
z = the z-score
σ = the standard deviation of the distribution
For the lower range:
x1 = 72 + (-1.96 * 3.5)
And for the upper range:
x2 = 72 + (1.96 * 3.5)
3. Calculate the values using the formula:
x1 = 72 - (1.96 * 3.5)
x2 = 72 + (1.96 * 3.5)
x1 ≈ 64.14
x2 ≈ 79.86
Therefore, in 95% of the years, the precipitation in this city is between approximately 64.14 and 79.86 inches.