Question

b. x is a normally distributed random variable with a standard deviation of 3.00. find the mean of x when 76.73% of the area lies to the left of 21.50.
c. x is a normally distributed random variable with a standard deviation of 14.00. find the mean of x when 82.38% of the area lies to the left of 86.

Explanation:

Step1: Find the z - score

For part b:
We know that if \(P(X < x)=0.7673\), using the standard normal table (z - table), the z - score \(z\) corresponding to an area of \(0.7673\) is \(z = 0.73\) (approximate value from the standard normal table).
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 21.50\), \(\sigma=3.00\) and \(z = 0.73\).
Substituting the values into the formula \(0.73=\frac{21.50-\mu}{3}\).

Step2: Solve for the mean \(\mu\)

Multiply both sides of the equation \(0.73=\frac{21.50 - \mu}{3}\) by \(3\):
\(0.73\times3=21.50-\mu\)
\(2.19 = 21.50-\mu\)
Then \(\mu=21.50 - 2.19\)
\(\mu = 19.31\)

For part c:
If \(P(X < x)=0.8238\), using the standard normal table (z - table), the z - score \(z\) corresponding to an area of \(0.8238\) is \(z = 0.93\) (approximate value from the standard normal table).
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 86\), \(\sigma = 14.00\) and \(z=0.93\)
Substitute the values into the formula \(0.93=\frac{86-\mu}{14}\)

Step3: Solve for the mean \(\mu\)

Multiply both sides of the equation \(0.93=\frac{86-\mu}{14}\) by \(14\):
\(0.93\times14=86-\mu\)
\(13.02=86-\mu\)
Then \(\mu=86 - 13.02\)
\(\mu=72.98\)

Answer:

For part b, the mean \(\mu = 19.31\). For part c, the mean \(\mu=72.98\)