Answer
a. From the given normal distribution, μ=2850 and σ= 420.
$P(x \lt 1200)$ = $P(z \lt \frac{1200-2850}{420})$
= $P(z \lt -3.9286)$
=.0 approximately.
b. $P(2300 \lt x \lt 3140)$ = $P(\frac{2300-2850}{420} \lt z \lt \frac{3140-2850}{420})$
= $P(-1.3095 \lt z \lt 0.6905)$
=0.7551 – 0.0952
=.6599 approximately.
c. $P(x \gt 3600)$ = $P(z \gt \frac{3600-2850}{420} )$
= $P( z \gt 0.3571)$
=0.9629
d. $P(3200 \lt x \lt 3700)$ = $P(\frac{3200-2850}{420} \lt z \lt \frac{3700-2850}{420})$
= $P(0.8333 \lt z \lt 2.0238)$
= 0. 9785 - 0.7977
=0.1808
Work Step by Step
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