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Nimfa-mama [501]
3 years ago
6

If Susie is late for school one particular day of the week, the probability she arrives on time the next day is 0.8. If she arri

ves on time one day, the probability she arrives late the following day is 0.3. Susie was on time on Monday. What is the probability she was late on Thursday
Mathematics
1 answer:
Andre45 [30]3 years ago
4 0

Answer:

0.273 = 27.3% probability that she was late on Thursday.

Step-by-step explanation:

On time on Monday, late on Thursday:

Possible outcomes (tuesday-wednesday-thursday).

on time(0.7 probability) - on time(0.7 probability) - late(0.3 probability)

on time(0.7 probability) - late(0.3 probability) - late(0.2 probability)

late(0.3 probability) - on time(0.8 probability) - late(0.3 probability)

late(0.3 probability) - late(0.2 probability) - late(0.2 probability).

What is the probability she was late on Thursday?

Sum of these four outcomes. So

p = 0.7*0.7*0.3 + 0.7*0.3*0.2 + 0.3*0.8*0.3+0.3*0.2*0.2 = 0.273

0.273 = 27.3% probability that she was late on Thursday.

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Given mean of 15.5 minutes , standard deviation of 1.7 minutes, sample size of 90 and sample mean of 15.4 minutes.

We can do the following study for conclusion:

Firstly the null hypothesis is

H_{0}: x=15.5

The alternate hypothesis is

H_{1}: x < 15.5

since the value is less than this is a one tailed test.

Z=x bar-x/d/\sqrt{n}

where x is sample mean and d is standard deviation.

Z=15.4-15.5/1.7/\sqrt{90}

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Critical value of Z at 0.1 level of significance

Z=-1.28

We fails to reject the null hypothesis since -0.560>-1,28

Hence we don't have evidence to support that the mean completion time under new management has decreased but we can conclude that it remains at 15.5 minutes.

Learn more about hypothesis at brainly.com/question/11555274

#SPJ4

Question is incomplete. It should include:

mean of 15.5 minutes ,

standard deviation of 1.7 minutes,

sample size of 90

and sample mean of 15.4 minutes.

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2 years ago
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