Question

Given a normal distribution with mean of 100 and standard deviation of 10, what is the probability that:

a. X > 80?

b. X < 65?

c. X < 75 or X > 90?

d. Between what two X values (symmetrically distributed around the mean) are ninety percent of the values

Answer 1

a.

X ~ N(100, 10²)

P(X > 80) = P[Z > (80 - 100)/10] = P[Z > -2] = 0.9772

b.

P(X < 65) = P[Z < (65 - 100)/10] = P[Z < -3.5] = 0.0002

c.

P(X < 75 or X > 90) = P(X < 75) + P(X > 90)

= P[Z < (75 - 100)/10] + P[Z > (90 - 100)/10]

= P[Z < -2.5] + P[Z > -1]

= 0.0062 + 0.8413

= 0.8475

d.

For ninety percent of the values, the percentiles are (1 - 0.9)/2 and 0.9 + (1 - 0.9)/2 which are  0.05 and 0.95

Z value for p = 0.05 and 0.95 are -1.645 and 1.645

The two values are,

100 - 1.645 * 10 = 83.55

100 + 1.645 * 10 = 116.45

Between 83.55 and 116.45, 90% of the data lies.