Question

Let X have a normal distribution with a mean of 23 and a standard deviation of 5. Find P(X < 9 or X > 21) in the following steps

(a) What region of the normal distribution are you looking to find the area of? (to the left of a zscore, to the right of a z-score, between two z-scores, or to the left of one z-score and to the right of another z-score)

(b) Calculate the z-score(s) needed to find the probability

(c) Find the probability P(X < 9 or X > 21)

Answer #1

Given that, mean (μ) = 23 and standard deviation = 5

We want to find, P(X < 9 or X > 21)

a) The region of the normal distribution is **to the left
of one z-score and to the right of another z-score.**

Because, (X < 9) shows the area to left of z-score and (X > 21) shows the area to the right of a z-score.

b) Z-score for x = 9 is,

Z = (9 - 23) / 5 = **-2.8**

Z-score for x = 21 is,

Z = (21 - 23) / 5 = **-0.4**

c) P(X < 9 or X > 21)

= P(X < 9) + P(X > 21)

= P(Z < -2.8) + P(Z > -0.4)

= [ 1 - P(Z < 2.8) ] + P(Z < 0.4)

= [ 1 - 0.9974 ] + 0.6554

= 0.0026 + 0.6554

= 0.6580

=> P(X < 9 or X > 21) = **0.6580**

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