Assume the random variable X is normally distributed with mean and standard deviation . Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. P(X)

The answer is attached for better presentation of formulas.

Step-by-step explanation:

The areas under the normal curve between any two ordinates at X=a and X=b equals the probability that the r.v X lies in the interval [a,b]. that is  P (a ≤ X ≤ b) = ∫_a^b▒1/(σ√2π) e^((-(x-u))/(2σ^2))  dx

which is represented by the area of the shaded region. (figure1)

But integrals of this type cannot be solved by ordinary means. They are however evaluated by the methods of numerical integration, and numerical approximations for some function have been tabulated for quick reference.

The table of areas under the unit normal curve gives the areas (probabilities) for the standard normal distribution from the mean, z=0 to a specified value of z say z0. Since normal curves are symmetrical therefore P (0 to z) = P (0 to –z). That is why the areas for negative values of z are not tabulated. Hence to use the table of areas for the normal distribution, the values of the r.v X in any problem are changed to the values of the standard normal variable Z and the desired probabilities are obtained from the table.

Thus to find P (a <X<b) we would change X into Z as follows

P (a <X<b) = P ((a-u)/σ ≤ (X-u)/σ  ≤  (b-u)/σ )

= P ((a-u)/σ ≤ Z ≤  (b-u)/σ )

Where (a-u)/σ and    (b-u)/σ  are the z- values of the standard normal variable Z.

In practice, a normal curve sketch for the given problem, showing under the X scale, a scale for the corresponding values of z will help in solving problem.  (Figure 2)