Answer
a. From the given normal distribution, μ= 25 and σ= 6 .
For x =29 : z = $\frac{29-25}{6}$ = $\frac{4}{6}$ = 0.67
For x =36 : z = $\frac{36-25}{6}$ = $\frac{11}{6}$ = 1.83
As shown in the figure, the required area is given by the area under the standard normal distribution curve between z=0.67 and 1.83 . This area is approximately 0.2191.
P($22 \leq x \leq 35$) = P( $0.67\leq z \leq 1.83$ ) = .6437 approximately.
b. For x =22 : z = $\frac{22-25}{6}$ = $\frac{4}{6}$ = -0.5
For x =35 : z = $\frac{35-25}{6}$ = $\frac{11}{6}$ = 1.67
As shown in the figure, the required area is given by the area under the standard normal distribution curve between z=-0.5 and 1.67 . This area is approximately 0.6437.
$P(22 \leq x \leq 35)$ = $P(-0.5 \leq z \leq 1.67 )$ = .6437 approximately.
Work Step by Step
See above
