Mathematics, 01.07.2020 15:01, bigmouth804

During the 2013 Major League Baseball season, the St. Louis Cardinals averaged 41,602 fans per home game. Suppose attendance during the season follows the normal probability distribution with a standard deviation of 9,440 per game. 95% of the attendance around the average is between (3,282", 22,722", 3,842",18,002"] and [69,922", 65,202", 60,482,", 79,362"] .

Answers: 1

Show answers

Answers

answered: alexabdercmur

About "95% of the attendance around the average is between" 22722 and 60482.

Step-by-step explanation:

The key to answering this question is the application of the 68-95-99.7 rule. It is a rule that "tells us" that data that follow a normal distribution approximately:

68% (68.27% to be more precise) of the observations are one standard deviation above and below the mean. We can express this mathematically as:

$\\ P(\mu - 1\sigma \leq X \leq \mu + 1\sigma) = 0.6827 \approx 0.68$ [1].

95% (95.45%) of the data are two standard deviations above and below the mean.

$\\ P(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545 \approx 0.95$ [2].

99.7% (99.73%) of the data are three standard deviations above and below the mean.

$\\ P(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973 \approx 99.7$ [3].

The two parameters that characterized the normal distribution are the mean, $\\ \mu$ , and the standard deviation, $\\ \sigma$ .

In this case, we have:

$\\ \mu = 41602$ . $\\ \sigma = 9440$ .

According to the 68-95-99.7 rule above explained, we know that the "95% of the attendance around the average" is two standard deviations, $\\ \sigma$ , above and below the mean, $\\ \mu$ .

Then, using [2], we have:

$\\ P(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545 \approx 0.95$ [2].

$\\ P(41602 - 2*9440 \leq X \leq 41602 + 2*9440) = 0.9545 \approx 0.95$ .

$\\ P(41602 - 18880 \leq X \leq 41602 + 18880) = 0.9545 \approx 0.95$ .

$\\ P(22722 \leq X \leq 60482) = 0.9545 \approx 0.95$ .

Therefore, about "95% of the attendance around the average is between" 22722 and 60482.

Notice that we approximate the values using the 68-95-99.7 rule, which is a good approximation and sufficient for practical uses. However, if we want to know exactly the 95% of the attendance around the average, we can calculate it using the formula for z-score (which, roughly speaking, "tell us" the distance from the mean in standard deviations units of a normally distributed value):

$\\ z = \frac{x - \mu}{\sigma}$ [4]

A positive value of z indicates that x is above the mean and a negative that x is below the mean.

Since the normal distribution is symmetrical around the mean, we can divide the 95% of probability in two equal parts. One above the mean, $\\ \frac{0.95}{2} = 0.475$ , and the other below the mean $\\ \frac{0.95}{2} = 0.475$

If we consult the cumulative standard normal table from the mean (0 to Z), the value for z, in this case, is $\\ z = 1.96$ , because for a probability of 0.475, the value of z = 1.96. Remember that the standard normal normal distribution has a $\\ \mu = 0$ and $\\ \sigma = 1$ . That is why the interval is from 0 (mean) to the value of Z.

Therefore, solving [4] for x, for an exactly value of 95% of the attendance around the average is between:

Upper value

The value is above the mean:

$\\ 1.96 = \frac{x - 41602}{9440}$

$\\ (1.96 * 9440) + 41602 = x$

$\\ x = (1.96 * 9440) + 41602$

$\\ x = 60104.40 \approx 60104$

Lower value

The value is below the mean:

$\\ -1.96 = \frac{x - 41602}{9440}$

$\\ x = (-1.96 * 9440) + 41602$

$\\ x = 23099.60 \approx 23100$

Or an exactly value of 95% of the attendance around the average is between 23100 and 60104. Notice that the question gives us options using the 68-95-99.7 rule, that is, 22722 and 60482, which are also valid (with less precision).

The below graphic shows the previous result (which is very similar to that obtained using the 68-95-99.7 rule.)