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"] .

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Answer from: 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.)

During the 2013 Major League Baseball season, the St. Louis Cardinals averaged 41,602 fans per home

Answer from: Quest

A=bxh3.2x1.8=5.76 centimeters square

Answer from: Quest

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation:

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