Answer
a. From the given normal distribution, μ= 65 and σ= 15 .
For x =45 : z = $\frac{45-65}{15}$ = -1.33
As shown in the figure, the required area is given by the area under the standard normal distribution curve to the left of z= -1.33. This area is 0.0912 approximately .
$P(x \lt 45)$ = $P(z \lt -1.33)$ = 0.0912 approximately
b. For x =79 : z = $\frac{79-65}{15}$ = 0.9333
As shown in the figure, the required area is given by the area under the standard normal distribution curve to the right of z= .9333. This area is approximately 0.1753.
$P(x \gt79)$ = $P(z \gt .9333 )$ = .1753 approximately
c. For x =54 : z = $\frac{54-65}{7}$ = -0.7333
As shown in the figure, the required area is given by the area under the standard normal distribution curve to the left of z= -0.7333 . This area is approximately 0.1753 .
$P(x \gt 54)$ =$ P(z \gt -0.7333)$ = .1753 approximately
d. For x =70 : z = $\frac{70-65}{15}$ = 0.3333
As shown in the figure, the required area is given by the area under the standard normal distribution curve to the left of z=0.3333 . This area is .6306 approximately.
$P(x \lt 70)$ =$ P(z \lt 0.3333)$ = .6306 approximately
Work Step by Step
see above
