## Problem 2001

Assume the SAT score are normally distributed with mean 1518 and standard deviation 325. Suppose you randomly select 100 SAT scores and want to find the probability that they have a mean less than 1500. How would you calculate the z score this situation?

Solution:-

Since you are selecting a group of 100 score, you would use the z-score formula for working with a group

z =

z =

z =

## Problem 1188

A binomial experiment is given. Decide whether you can the normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why.

A  survey a adult found that 58% have used a multivitamin in the past 12 months. You randomly select 55 adults and ask them if they have used a multivitamin in the past 12 months.

Solution:-

If np ≥ 5 and np ≥ 5, then the binomial random variable, x is approximately normally distributed, with mean µ = np and standard deviation σ = , where n is the sample size, p is the population proportion, and q = 1 – p.

Find np and nq to determine if the normal distribution can be used to approximate the binomial distribution. First determine the values of  n, p, and q.

n = 55

p = = 0.58

q = 1 – 0.58 = 0.42

Now calculate np and nq.

np = (55)(0.58) = 31.9

nq = (55)(0.42) = 23.1

Since both np and nq are greater than 5, the normal distribution can be used to approximate the binomial distribution. Thus, calculate the mean, µ = np, and standard deviation, σ = . Recall that up = 31.9. Find the standard deviation, rounding to two decimal places.

σ =

Therefore, the mean is 31.9 and the standard deviation is approximately 3.66.

## Problem 1082

Find the equation of the parabola determine by the given information.

Focus (6,0), directrix x = -6

Solution:-

First, notice that since the directrix is a vertical line, the parabola will open either to the left or to the right, not up or down. Since the directrix is to the left of the focus, and a parabola always opens away from its directrix, this parabola will open to the right.

The vertex will be halfway between the focus and the directrix. It will have the same y-coordinate as the y-coordinate of the focus.

The vertex of the parabola is , which simplifies to (0,0).

The parabola with directrix x = -p and vertex (0,0) will have the equation y2 = 4px.

Since the directrix of this parabola is x = -6 , p = 6.

The equation of the parabola with directrix x = -6 and focus (6,0) is  y2 = 4*6x, which simplify to y2 = 24x.

## Problem 1081

Find the vertex, the focus, and the directrix.  Then draw the graph.

y2 = -6x

Solution:-

An equation of the form y2 = 4px is the standard equation of a parabola with the vertex at the origin, focus at (p,0) and directrix x = -p.

To write the equation y2 = -6x in the form y2 = 4px, p = –

Thus , y2 = -6x is equivalent to y2 = 4(-)x.

The vertex is located at (0,0).

The focus is at (-,0).

The directrix is the line x=.

The correct graph of the relation is shown to the right.

## Problem 1047

Indicate whether the statement is a simple or a compound statement. If it is a compound statement, indicate whether it is a negation, conjunction, dis-junction, conditional, or bi-conditional by using both the word and its appropriate symbol.

Statement: If it is raining outside, then we will stay inside.

Solution:-

Sentences that convey only one idea are called simple statement. Statements consisting of two or more simple statements are called compound statements.

A negation change a statement to its opposite meaning. A conjunction is an ‘and’ statement, and a dis-junction is an ‘or’ statement. A conditional is an ‘if-than’ statement, and a bi-conditional is an ‘if and only if’ statement.

There are two simple statement contained in the statement above: “it is raining outside,” “and “we will stay inside.”

Since there are two simple statement contained in the statement, it is a compound statement.

The “If….then” words in the statement indicate that it is a conditional compound statement.

## Problem 1042

The birth weights of full-term babies are normally distributed with mean 3800 grams and standard deviation σ = 470 grams. Use this information to do the following.

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

(b)Shaded the region that represents the proportion of full-term babies who weigh more than 4740 grams.

(c)   Suppose the area under the normal curve to the right of X = 4740 is 0.0228. 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 = 38000 – 470 = 3330

µ +σ = 3800 + 470 = 4270

The graph that is symmetrical about 3800 and has inflection points at 3330 and 4270 is shown to the right.

(b)  Shade the region that represents the proportion of full-term babies who weigh more than 4710 grams.

For the given normal distribution curve, the area under the curve to the left of a value shows the percentage of births that weigh less than that value, and the area under the curve to the right of a  value shows the percentage of births that weigh more than that value.

Since we wish to shade the region that shown the full-term babies that weigh more than 4740 grams, we need to shade to the right of 4740.

The graph that shown the region to the right of 4740 shaded is shown to the right. Notice that 4740 is two standard deviations away from the mean.

(c ) Suppose the area under the normal curve to the right of X = 4740 is 0.0228. Provide an interpretation of this result.

Remember that for the given normal distribution curve, the area under to the left of a value shows the percentage of births that weigh less than that value, and the area under the curve to the right of a value shows the percentage of births that weigh more than that value.

Since the area under the normal curve to the right of X = 4710 represents the percentage of full-term babies that weigh more than 4740 grams. 2.28% of full-term babies weigh more than 4740 grams, or the probability is 0.0228 that a randomly selected full-term baby weighs more than 4740 grams.

## 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.

(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. (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 1040

One graph in the figure represents a normal distribution with mean µ = 12 and standard deviation σ = 3.

The other graph represents a normal distribution with mean µ = 7 and standard deviation σ = 3.

Determine which graph is which and explain how you know.

Solution:-

Graph A has a mean of µ = 7 and graph B has a mean of µ = 12 because a larger mean shift the graph to the right.

## Problem 1039

The relative frequency histogram represents the length of phone calls on George’s cell phone during the month of September. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.

Solution:-

No, because the histogram does not have the shape of a normal curve.

## Problem 1038

A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.

Solution:-

The histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable.