Exercise 1 Page 806

Practice makes perfect

a Let

X

represent the number of uses of the GPS system by a random user. We want to find the probability that a randomly chosen user will have used the GPS more than

20

times, which is the same as finding the probability that the value of our random variable

X

is greater than

20 .

P (X > 20) = ?

Before we start calculating let's look at the given table.

Uses	Customers
$0$	$1382$
$1 - 5$	$2350$
$6 - 10$	$2010$
$11 - 15$	$1863$
$16 - 20$	$1925$
$21 +$	$470$

Since $X$ represents the number of uses of a GPS, we can rewrite the first column using our random variable $X .$

Value of $X$	Customers
$X = 0$	$1382$
$1 \leq X \leq 5$	$2350$
$6 \leq X \leq 10$	$2010$
$11 \leq X \leq 15$	$1863$
$16 \leq X \leq 20$	$1925$
$X \geq 21$	$470$

To find

P (X > 20)

we need to find the number of customers that have used the GPS system more than

20

times. Looking at the table we can tell that there is only one outcome where

X > 20 .

To find the probability, we will use the Probability Formula.

P (X > 20) = \frac{Customers for whom X > 2 0}{Customers surveyed}

We are given that the car dealership surveyed

10000

of its customers with a GPS and we have found that the number of customers for whom

X > 20

470 .

P (X > 20) = \frac{4 7 0}{1 0 0 0 0} = 0.047

The probability that a randomly chosen customer will have used the GPS system more than

20

times is equal to

0.047,

which can also be written as

4.7 % .

b This time we want to find the probability that a randomly chosen customer will have used the GPS system no more than

10

times. This is the same as finding the probability that the value of our random variable

X

is less than or equal to

10 .

P (X \leq 10) = ?

Let's look at the table we made in Part A again.

Value of $X$	Customers
$X = 0$	$1382$
$1 \leq X \leq 5$	$2350$
$6 \leq X \leq 10$	$2010$
$11 \leq X \leq 15$	$1863$
$16 \leq X \leq 20$	$1925$
$X \geq 21$	$470$

We will consider the first

3

rows of our table only because the value of

X

is less than or equal to

10

in these rows. Our event is made up of

3

more simple events, so this is a compound event. Since the

3

events are mutually exclusive events, we can calculate

P (X \leq 10)

by adding the probabilities that each event occurs.

P (X \leq 10) P (X = 0) + P (1 \leq X \leq 5) = + P (6 \leq X \leq 10)

We will calculate each probability using the Probability Formula. Probability of each event is equal to the number of customers that are related to the given event divided by the total number of customers which is

10000 .

Value of $X$	Customers	Probability	Simplified
$X = 0$	$1382$	$\frac{1 3 8 2}{1 0 0 0 0}$	$0.1382$
$1 \leq X \leq 5$	$2350$	$\frac{2 3 5 0}{1 0 0 0 0}$	$0.235$
$6 \leq X \leq 10$	$2010$	$\frac{2 0 1 0}{1 0 0 0 0}$	$0.201$

Now we can substitute in the values from the table into our equation for

P (X \leq 10) .

P (X \leq 10) 0.1382 + 0.235 + 0.201 = = 0.5742 \approx 0.574

The probability that a randomly chosen customer will have used the GPS system no more than

10

times is equal to

0.574,

which can also be written as

57.4 % .

Subchapter links

Check Your Understanding p.806

Practice and Problem Solving p.806-808

H.O.T. Problems p.808

Standardized Test Practice p.809

Spiral Review p.809

Exercise 1 Page 806

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