A variable is normally distributed with mean 68 and standard deviation 10. Find:
a) The value that 85% of all possible values of the variable exceed.
b) The two values that divide the area under the corresponding normal curve into a middle area of 0.90 and two outside areas of 0.05.
To find the values in the normal distribution, we can make use of the Z-score formula:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the value in the distribution
μ is the mean of the distribution
σ is the standard deviation of the distribution
Given a normal distribution with mean μ = 68 and standard deviation σ = 10, we can solve for the value X using the Z-score formula.
a) To find the value that 85% of all possible values exceed, we need to determine the Z-score for the corresponding area under the curve.
The area under the normal curve is divided into two equal tails since we want to find the value that 85% of all possible values exceed. This means that we have an area of 100% - 85% = 15% in the tail.
To find the Z-score associated with an area of 15% in the tail, we can refer to the standard normal distribution table or use a calculator that provides Z-score calculations. For simplicity, let's consult the standard normal distribution table.
The standard normal distribution table provides the area to the left of the Z-score. Since we want to find the area in the tail (to the right), we will need to subtract the given tail area from 100%.
Looking up for 0.15 (15%) in the standard normal distribution table, we find the Z-score of approximately 1.0364.
Now we can substitute the known values into the Z-score formula and solve for X:
1.0364 = (X - 68) / 10
Rearranging the equation and solving for X:
X - 68 = 1.0364 * 10
X - 68 = 10.364
X = 78.364
Therefore, the value that 85% of all possible values exceed is approximately 78.364.
b) To find the two values that divide the area under the corresponding normal curve into a middle area of 0.90 and two outside areas of 0.05 each, we will first determine the Z-scores associated with the given tail areas.
For the two outside areas of 0.05 each, we divide 0.05 by 2 to get 0.025. We need to find the Z-scores that correspond to an area of 0.025 in each tail.
Looking up for 0.025 (2.5%) in the standard normal distribution table, we find the Z-score of approximately -1.9599.
Now, to find the Z-score that corresponds to the middle portion of 0.90 (or 90%), we subtract the sum of the two tail areas from 1:
Middle area = 1 - (2 * tail area)
Middle area = 1 - (2 * 0.05)
Middle area = 1 - 0.1
Middle area = 0.9
Since we know the area to the left of the Z-score, we need to find the Z-score corresponding to an area of 0.9 in the standard normal distribution table. Looking up for 0.9 (90%), we find the Z-score of approximately 1.2816.
Now we substitute the Z-scores into the Z-score formula and solve for X:
For the left value:
-1.9599 = (X - 68) / 10
X - 68 = -1.9599 * 10
X - 68 = -19.599
X = 48.401
For the right value:
1.2816 = (X - 68) / 10
X - 68 = 1.2816 * 10
X - 68 = 12.816
X = 80.816
Therefore, the two values that divide the area under the corresponding normal curve into a middle area of 0.90 and two outside areas of 0.05 each are approximately 48.401 and 80.816.