Question

suppose the monthly charges for cell phone plans are normally distributed with mean \\( \mu = \\$ 66 \\) and standard deviation \\( \sigma = \\$ 17 \\).
(a) draw a normal curve with the parameters labeled.
(b) shade the region that represents the proportion of plans that charge less than \\( \\$ 49 \\).
(c) suppose the area under the normal curve to the left of \\( x = \\$ 49 \\) is \\( 0.1587 \\). provide an interpretation of this result.
(c) select the correct choice below and fill in the answer box to complete your choice.
(type a whole number.)
a. the probability is \\( 0.1587 \\) that a randomly selected cell phone plan in this population is less than \\( \\$ \\) per month.
b. the probability is \\( 0.1587 \\) that a randomly selected cell phone plan in this population is more than \\( \\$ \\) per month.

Explanation:

Step1: Analyze the normal distribution parameters

Given the normal distribution with mean \(\mu = 66\) and \(X = 49\). The region representing the proportion of plans that charge less than \(49\) should be to the left of \(X = 49\) on the normal curve. Looking at the options for part (b), option C has the shaded region to the left of \(49\) (since in a normal - distribution graph, the value \(49\) is less than the mean \(66\), and the left - hand side of \(49\) relative to the mean is the correct shaded area).

Step2: Calculate the probability for part (c)

We know that the total area under the normal curve is \(1\). If \(P(X\lt49)=0.1587\), then \(P(X > 49)=1 - P(X\lt49)\).

$$P(X>49)=1 - 0.1587=0.8413$$

Answer:

(b) C; (c) \(0.8413\)