## Related questions with answers

The number of tickets sold at high school basketball games in a particular conference are normally distributed with a mean of 68.7 and a standard deviation of 13.1.

About what percent of the games sell fewer than 75 tickets?

Solution

VerifiedGiven:

$\begin{align*} \mu&=\text{Mean}=68.7 \\ \sigma&=\text{Standard deviation}=13.1 \\ x&=75 \end{align*}$

The z-score is the value decreased by the mean, divided by the standard deviation.

$z=\dfrac{x-\mu}{\sigma}=\dfrac{75-68.7}{13.1}=0.48$

Determine the corresponding probability using the normal probability table in the appendix. $P(Z<0.48)$ is given in the row starting with $0.4$ and in the column starting with .08 of the standard normal probability table in the appendix.

$P(X<75)=P(Z<0.48)=0.6844=68.44\%$

Thus about 68.44% of the games sell fewer than 75 tickets.

$\text{\color{#4257b2}TI83/84-calculator command: normalcdf(-1E99, 0.48), where -1E99 can be replaced by any other very small number (-1E99 is the smallest number accepted by the calculator).}$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Algebra 2 Common Core Edition

1st Edition•ISBN: 9780076639908Carter, Casey, Cuevas, Day, Holliday, Malloy#### Glencoe Algebra 2

1st Edition•ISBN: 9780079039903Gilbert J. Cuevas, John A. Carter, Roger Day#### Algebra 2 Common Core

1st Edition•ISBN: 9780133186024 (4 more)Basia Hall, Charles, Kennedy#### Big Ideas Math Algebra 2: A Common Core Curriculum

1st Edition•ISBN: 9781608408405 (1 more)Boswell, Larson## More related questions

1/4

1/7