Find the number x such that the proportion of observations that are less than x in a Normal distribution with mean 100 and standard deviation 25 is 0.3. (Use 2 decimal places.)
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To find the number x such that the proportion of observations that are less than x in a Normal distribution, you can use the z-score formula and then convert it back to the original scale.
The z-score formula is given by:
z = (x - μ) / σ
Where:
z is the z-score
x is the value we want to find
μ is the mean of the distribution
σ is the standard deviation of the distribution
In this case, the proportion of observations that are less than x is 0.3. This means that the area to the left of x under the normal curve is 0.3.
To find the z-score corresponding to this proportion, we can use a standard normal table or a statistical calculator. From the standard normal table, we find that the z-score corresponding to 0.3 is approximately -0.5244.
Now, we can rearrange the z-score formula to solve for x:
z = (x - μ) / σ
-0.5244 = (x - 100) / 25
Solving for x, we have:
-0.5244 * 25 = x - 100
-13.11 = x - 100
x = 100 - 13.11
x ≈ 86.89
Therefore, the number x such that the proportion of observations that are less than x in a Normal distribution with mean 100 and standard deviation 25 is 0.3 is approximately 86.89.