normal distribution

**m777** · June 1st 2007, 05:10 AM

Hello,
plz try to solve this question.
Thanks.
Question:
Given a normal distribution with μ=40 and σ = 6 , find
(a) the area below 32
(b) the area above 27
(c) the area between 42 and 51
(d) the x value that has 45% of the area below it
(e) the x value that has 13% of the area above it

**ThePerfectHacker** · June 1st 2007, 06:15 AM

Originally Posted by m777

Given a normal distribution with μ=40 and σ = 6 , find
(a) the area below 32

Find the z-score,
$\frac{40-32}{6} = z \to z =1.33$
The corresponding value is .408 on the table.
Thus, $P(32 \leq x \leq 40) = .408$ .
But then,
$P(x\leq 32) + P(32\leq x \leq 40) = .5$
Thus,
$P(x\leq 32) = .092$

(b) the area above 27

Find the corresponding z-score:
$\frac{40-27}{6}\approx 2.166$
The corresponding value is .484
Thus,
$P(27\leq x\leq 40) = .484$
But then,
$P(27\leq x) = P(27\leq x \leq 40)+P(40\leq x) = .484+.5 = .984$

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