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The average time spent sleeping (in hours) for a group of medical residents at a hospital can be approximated by a normal distribution. The average is 6.1 hours with a standard deviation of 1 hour. a. What is the shortest time spent sleeping that would place a resident in the top 5% of sleeping times

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Answer:

7.75 hours

Step-by-step explanation:

Given that:

Mean, m = 6.1

Standard deviation, s = 1

Shortest Time spent sleeping in other to be in top 5% of sleeping time.

5% = 0.05

Zscore for P(Z > 0.05) = 1.645 (Z probability calculator)

To obtain the shortest time, x

Zscore = (x - m) / s

1.645 = (x - 6.1) / 1

1.645 = x - 6.1

1.645 + 6.1 = x

7.745 = x

x = 7.75

Sgirtws sleep time = 7.75 hours

The shortest time spent sleeping which would place a resident in the top 5% of sleeping times is 7.75 hours.

How to get the z scores?

The value of z score can be find out using the following formula.

[tex]Z = \dfrac{X - \mu}{\sigma}[/tex]

Here, (X) is the mean of the sample, (σ) is the standard daviation and (μ) is the mean.

The average time spent sleeping (in hours) for a group of medical residents at a hospital can be approximated by a normal distribution.

The average is 6.1 hours with a standard deviation of 1 hour.

  • The mean (μ) of the data is 6.1
  • The standard deviation (σ) is 1.

The shortest time spent sleeping that would place a resident in the top 5% of sleeping times has to be find out. This shortest time is mean of the sample (X).

The level of significance is 0.05 and the z score at this value will be 1.645. Put the values in the above formula,

[tex]1.645 = \dfrac{X - 6.1}{1}\\X\approx7.75[/tex]

Thus, the shortest time spent sleeping which would place a resident in the top 5% of sleeping times is 7.75 hours.

Learn more about the z score here;

https://brainly.com/question/13299273

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