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denpristay [2]
2 years ago
15

A basketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. She misses the first shot 40% o

f the time. When she misses the first shot, she misses the second shot 5% of the time. What is the probability of missing both free-throw shots?
Mathematics
1 answer:
yan [13]2 years ago
5 0
<span>The probability of missing the first shot is 40%. When she misses the first shot, the probability of missing the second shot is very, very low, only 5%. That means that the probability of missing both shots must be much smaller than the probability of missing the first shot.
 
40/100 * 5/100
0.4 * 0.05
= 0.02
= 2%</span>
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Previous concepts

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\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

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And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

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2 years ago
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