Posts Tagged ‘normally distributed’

Problem 1041

Suppose the monthly charge for cell phone plans is normally distributed with mean µ = $75 and standard deviation σ = $21.

(a)    Draw a normal curve with the parameters labeled.

(b)   Shade the  region that represents the proportion of plans that charge less than $54.

(c)  Suppose the area under the normal curve to the left of X = $54 is 0.1587. Provide an interpretation of this result.

 

Solution:-

 

(a)  The graph of a normal curve is symmetric about the mean, µ, and has inflection points at µ ± σ.

Determine the location of the  inflection points.

µ – σ = 75 – 21 = 54

µ +σ = 75 + 21 = 96

 

The graph that is symmetrical about 75 and  has inflection points at 54 and 96 is shown to the right.

 

17c

(b)  For the given normal distribution curve, the area under to the left of a value shows the percentage of cell phone plans that cost than value, and the area under the curve to the right of a value shows the percentage of cell phone that cost more than the value.

Since we wish to shade the region that shows the proportion of plans that cost less than $54, we need to shade to the lest of 54.

 

The graph that shows the region to left of 54 shaded is shown to the right.

17d

 

(c)  Remember that for the given distribution curve, the area under the curve to the left of a value shows the percentage of cell phone plans that cost less than that value, and the area under the curve to the right of a value shows the percentage of cell phone plans that cost more than that value.

 

Since the area under the normal curve to the left of X = $54 represents the percentage of plans that cost less than $54, 15.87% of cell phone plans cost less than $54 per month, or the probability is 0.1587 that a randomly selected cell phone plan costs plan costs less than $54 per month.

 

Problem 651

The time required to finish a test is normally distributed with a mean of 60 minutes and a standard deviation of 10 minutes. What is the probability that a student will finish the test in less than 70 minutes?

 

Solution:-

84%

Problem 650

The time required to finish a test is normally distributed with a mean of 60 minutes and a standard deviation of 10 minutes. What is the z-Score for a student who finishes the test in 45 minutes?

 

Solution:-

 

z = \frac{(45 - 60)}{10}

z =\frac{ -15}{10}

z = -1.5