Answer:
the third one
Step-by-step explanation:
<h2>
Answer with explanation:</h2>
Let be the population mean lifetime of circulated $1 bills.
By considering the given information , we have :-
Since the alternative hypotheses is two tailed so the test is a two tailed test.
We assume that the lifetime of circulated $1 bills is normally distributed.
Given : Sample size : n=50 , which is greater than 30 .
It means the sample is large so we use z-test.
Sample mean :
Standard deviation :
Test statistic for population mean :-
The p-value=
Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.
Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.
For 1 play, the chance of gaining $8 is 4/38, while the chance of losing the $1 is 34/38. Therefore, the expected value is ($8)(4/38) + ($-1)(34/38) = $(-1/19). Over 50 plays, which are mutually independent of each other, we multiply the number of plays by the expected value to get $(-50/19) = $-2.63.
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I have to do a lot of stuff to do this in the time of time I have to do a lot of work to do this really
(-9,1) should be the answer not completely sure tho