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dusya [7]
3 years ago
9

One of the events in the Winter Olympics is the Men's 500-meter Speed Skating.

Mathematics
1 answer:
poizon [28]3 years ago
8 0

Answer:

Mean: \bar x = 40.61

Median =40.2

Mode = 43.4

Step-by-step explanation:

Given

See attachment for data

Solving (a): The mean

Mean is calculated as:

\bar x = \frac{\sum x}{n}

From the attached:

n = 14

So, the mean is:

\bar x = \frac{43.4+43.4+43.4+43.1+43.2+40.2+40.2+40.1+40.3+39.44+39.17+38.03+38.19+36.45}{14}

\bar x = \frac{568.58}{14}

\bar x = 40.61

Solving (b): The median

n = 14; this is an even number. So, the median is:

Median = \frac{1}{2}(n+1)

Median = \frac{1}{2}(14+1)

Median = \frac{1}{2}(15)

Median = 7.5th

This implies that the median is the average of the 7th and 8th item.

Next, is to order the data (in ascending order): <em>36.45, 38.03, 38.19, 39.17, 39.44, 40.1, 40.2, 40.2, 40.3, 43.1, 43.2, 43.4, 43.4, 43.4.</em>

The 7th and 8th items are: 40.2 and 40.2

The median is:

Median = \frac{1}{2}(40.2 + 40.2)

Median = \frac{1}{2}*80.4

Median =40.2

Solving (c): The mode

43.4 has the highest number of occurrence.

So:

Mode = 43.4

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Answer:

The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

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P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0

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Based on the Central Limit Theorem:

\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)

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the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

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