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3 years ago

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Assume the population of regulation basketball weights are normally distributed with a mean of 22 and a standard deviation of 1

ounce. If a sample of 100 regulation basketballs is taken, what is the probability that its sample mean will be greater than 22.2 ounces

1 answer:

ankoles [38]3 years ago

6 0

Answer: 0.0228

Step-by-step explanation:

please check photo explanation

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

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1. Divide wire b in parts x and b-x.

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There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
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Another way is apply the formula A=1/2*base*altitude,
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5. f(x) is a minimum means f'(x)=0

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$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

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