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 sprint 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.

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%

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: