Question

# 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

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Given:

\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.

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

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