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The results of a national survey showed that on average, adults sleep 6.6 hours per night. Suppose that the standard deviatio

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Answer #1

u= average = 6.6

0 = standard deviation = 1

Empirical rule

1) Approrimately 68% of the data will be between u-o and u to

We plug the value of average and standard deviation

Approximately 68% of the data will be between p - 0 = 6.6 – 1 = 5.6 and uto= 6.6+1 = 7.6

Approximately 68% of the data will be between 5.6 and 7.6

2) Approximately 95% of the data will be between u-2*0 and u +2*0.

We plug the value of the average and standard deviation.

Approximately 95% of the data will be between u-2*0 = 6.6 – 2*1 = 6.6 – 2 = 4.6 and u +2*0= 6.6+2*1 = 6.6 +2 = 8.6

Approximately 95% of the data will be between 4.6 and 8.6

3) Approximately 99.7% of the data will be between u-3*o and u+3*0.

We plug the value of the average and standard deviation.

Approximately 99.7% of the data will be between u-3*0 = 6.6–3*1 = 6.6 – 3 = 3.6 and u +3* = 6.6+3*1= 6.6+3 = 9.6

Approximately 99.7% of the data will be between 3.6 and 9.6

Answer for # a part

Approximately 95% of the data will be between 4.6 and 8.6

We get the answer as 95 % for # a part

#b part

x: number of hours sleeps per night in b part = 8

u= average = 6.6

0 = standard deviation = 1

Formula for Z score

- Z score = -

Z score = 8 – 6.6 -= 1.4

We round the z score value up to 2 decimal place

Z score = 1.40

Final answer for b part

Z score = 1.40

# c part

x: number of hours sleeps per night in c part = 6

u= average = 6.6

0 = standard deviation = 1

Formula for Z score

- Z score = -

Z score = 6 – 6.6 - = -0.6

We round the z score value up to 2 decimal place

Z score = -0.60

Final answer for c part

Z score = - 0.60

I hope this will help you :)

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