Question

the data set shows the number of points students scored in a video game tournament:
\\{88, 94, 90, 84, 85, 80, 88, 87, 93, 86, 88, 95, 92, 80, 90\\}. what is the probability that a student scored at most 87 points in the video game tournament?

use the geogebra spreadsheet to find the mean and standard deviation. use these values to create a normal distribution for the data set. then use the geogebra probability calculator to solve the problem.

use the drop-down arrow to choose the correct word to complete the sentences.

about % of the student scored at most 87 points, with mean score of points and a standard deviation of points.

Explanation:

Calculate the sample mean

We first find the mean of the given data set:

$$ \{88, 94, 90, 84, 85, 80, 88, 87, 93, 86, 88, 95, 92, 80, 90\} $$

The total number of data points is \(n = 15\).
Using the Sample Mean concept:

$$ \mu = \frac{\sum x_i}{n} = \frac{88 + 94 + 90 + 84 + 85 + 80 + 88 + 87 + 93 + 86 + 88 + 95 + 92 + 80 + 90}{15} $$

$$ \mu = \frac{1316}{15} \approx 87.7333 $$

Calculate the sample standard deviation

Using the Sample Standard Deviation concept:

$$ s = \sqrt{\frac{\sum (x_i - \mu)^2}{n - 1}} $$

Let's compute the squared differences from the mean \(\mu \approx 87.7333\):

• \((88 - 87.7333)^2 \approx 0.0711\)
• \((94 - 87.7333)^2 \approx 39.2711\)
• \((90 - 87.7333)^2 \approx 5.1378\)
• \((84 - 87.7333)^2 \approx 13.9378\)
• \((85 - 87.7333)^2 \approx 7.4711\)
• \((80 - 87.7333)^2 \approx 59.8044\)
• \((88 - 87.7333)^2 \approx 0.0711\)
• \((87 - 87.7333)^2 \approx 0.5378\)
• \((93 - 87.7333)^2 \approx 27.7378\)
• \((86 - 87.7333)^2 \approx 3.0044\)
• \((88 - 87.7333)^2 \approx 0.0711\)
• \((95 - 87.7333)^2 \approx 52.8044\)
• \((92 - 87.7333)^2 \approx 18.2044\)
• \((80 - 87.7333)^2 \approx 59.8044\)
• \((90 - 87.7333)^2 \approx 5.1378\)

Sum of squared differences:

$$ \sum (x_i - \mu)^2 \approx 293.0667 $$

$$ s = \sqrt{\frac{293.0667}{14}} \approx \sqrt{20.9333} \approx 4.5753 $$

Rounding to the nearest tenth, the mean is \(87.7\) and the standard deviation is \(4.6\).

Calculate the normal distribution probability

Using the Normal Distribution Probability concept, we model the scores with a normal distribution \(X \sim N(\mu = 87.7333, \sigma = 4.5753)\).
We want to find the probability that a student scored at most 87 points, \(P(X \le 87)\):

$$ z = \frac{87 - 87.7333}{4.5753} \approx -0.1603 $$

Using a standard normal distribution table or calculator:

$$ P(Z \le -0.1603) \approx 0.4363 \approx 43.6\% $$

Rounding to the nearest whole percent, this is about \(44\%\).

Answer:

About <blank>44</blank>% of the student scored at most 87 points, with mean score of <blank>87.7</blank> points and a standard deviation of <blank>4.6</blank> points.