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alexandr1967 [171]
3 years ago
11

Bob and Doug are both 100-meter sprinters. Bob's sprint time is normally distributed with a mean of 10.00 seconds and Doug's spr

int time is also normally distributed, but with a mean of 9.90 seconds. Both have the same standard deviation in sprint time σ. Assuming that Bob and Doug have independent sprint times, and given that there is a .95 chance that Doug beats Bob in any given race, find σ.
Mathematics
1 answer:
alexdok [17]3 years ago
3 0

Answer:

The standard deviation (σ) = 0.05

Step-by-step explanation:

The question is to find the standard deviation.

STEEP 1: FIND THE MEAN

(10+9.9) ÷ 2 = 9.95

STEP 2: SQUARE THE DIFFERENCE BETWEEN SPRINT TIME AND MEAN

10-9.95= 0.05

0.05^2 = 0.0025

9.9 - 9.95= -0.0025

-0.0025^2 = 0.0025

STEP 3: FIND THE VARIANCE

0.0025+0.0025= 0.005

0.005/2= 0.0025

STEP 4: FIND THE STANDARD DEVIATION (σ )

√variance

√0.0025 = 0.05

Therefore

σ = 0.05.

From the standard deviation, the percentage probability of the higher value to occurs is

0.05×100= 5%

That means Doug has 95%

And Bob has 5%

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Step-by-step explanation:

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<u><em>Answer:</em></u>

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The three problems deal with inverse variation between two variables

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<u>Mathematically, an inverse variation relation is represented as follows:</u>

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<u><em>Now, let's check the givens:</em></u>

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We are given that y = 3 and k = 33

<u>Substitute in the original relation and solve for x as follows:</u>

y = \frac{k}{x}\\ \\3 = \frac{33}{x}\\ \\x=\frac{33}{3}=11

<u>Part b:</u>

We are given that y = 11 and x = 5.2

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<u>Part c:</u>

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