8) At a large spikeball tournament, the length of time of each rally, in seconds, is normally distributed with mean and standard deviation o. It is found that 4% of the rallies take fewer than 5 seconds, and 70% of the rallies take fewer than 20 seconds. Find the values of µ and o. Solution? Using the inverse normal function, we calculate the z-scores. We set the areas to 0.04 and 0.7. In both cases we let u = 0,0 = 1. The two z-scores are –1.750686 .. and 0.524400. = -1.750686 .. We set up a system of equations: 20-u = 0.524400 ... Solving using technology, we find µ = 6.59 and o = 16.5
8) At a large spikeball tournament, the length of time of each rally, in seconds, is normally distributed with mean and standard deviation o. It is found that 4% of the rallies take fewer than 5 seconds, and 70% of the rallies take fewer than 20 seconds. Find the values of µ and o. Solution? Using the inverse normal function, we calculate the z-scores. We set the areas to 0.04 and 0.7. In both cases we let u = 0,0 = 1. The two z-scores are –1.750686 .. and 0.524400. = -1.750686 .. We set up a system of equations: 20-u = 0.524400 ... Solving using technology, we find µ = 6.59 and o = 16.5
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![8) At a large spikeball tournament, the length of time of each rally, in seconds, is normally distributed with mean u
and standard deviation o. It is found that 4% of the rallies take fewer than 5 seconds, and 70% of the rallies take
fewer than 20 seconds. Find the values of µ and o.
Solution?
Using the inverse normal function, we calculate the z-scores.
We set the areas to 0.04 and 0.7. In both cases we let u = 0,0 = 1.
The two z-scores are – 1.750686 . and 0.524400 .
5-H
-1.750686 .
We set up a system of equations:
20-u
= 0.524400 ...
Solving using technology, we find u z 6.59 and o z 16.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c102743-b2a4-48f3-a0c1-996319469ac3%2F260ac897-5662-4256-a67d-0305183c450a%2Fqtni33_processed.png&w=3840&q=75)
Transcribed Image Text:8) At a large spikeball tournament, the length of time of each rally, in seconds, is normally distributed with mean u
and standard deviation o. It is found that 4% of the rallies take fewer than 5 seconds, and 70% of the rallies take
fewer than 20 seconds. Find the values of µ and o.
Solution?
Using the inverse normal function, we calculate the z-scores.
We set the areas to 0.04 and 0.7. In both cases we let u = 0,0 = 1.
The two z-scores are – 1.750686 . and 0.524400 .
5-H
-1.750686 .
We set up a system of equations:
20-u
= 0.524400 ...
Solving using technology, we find u z 6.59 and o z 16.5
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