Question

use technology to find the indicated area under the standard normal curve. include an appropriately labeled sketch of the normal curve and shade the appropriate region
a. find the probability that a z - score will be 0.91 or less
b. find the probability that a z - score will be 0.91 or more
c. find the probability that a z - score will be between - 1.4 and - 1.04
the probability that a z - score will be 0.91 or less is 0.8186
(round to four decimal places as needed.)
b. which graph below shows the probability that a z - score is 0.91 or more?

Explanation:

Step1: Use the property of the standard normal distribution

The total area under the standard normal curve is \(1\). For part (a), using a calculator or software (e.g., in R: pnorm(0.91) or in Excel: NORM.S.DIST(0.91,TRUE)), we find the cumulative probability.

Step2: For part (b)

The probability that \(Z\geq0.91\) is \(1 - P(Z < 0.91)\). Since \(P(Z < 0.91)=0.8186\) (from part (a)), then \(P(Z\geq0.91)=1 - 0.8186=0.1814\). The graph that shows the probability that a \(z -\)score is \(0.91\) or more is the one where the shaded region is to the right of \(z = 0.91\), which is option C.

Step3: For part (c)

Using the formula \(P(-1.4<Z<-1.04)=P(Z < - 1.04)-P(Z < - 1.4)\). In R, pnorm(-1.04)-pnorm(-1.4); in Excel, NORM.S.DIST(-1.04,TRUE)-NORM.S.DIST(-1.4,TRUE)

$$P(Z < - 1.04)=0.1492\quad\text{and}\quad P(Z < - 1.4)=0.0808$$

$$P(-1.4 < Z < - 1.04)=0.1492 - 0.0808=0.0684$$

Answer:

a. \(0.8186\)
b. Option C
c. \(0.0684\)