Question

A recent survey describes the total sleep time per night among college students as approximately Normally distributed with mean μ = 6.72 hours and standard deviation σ = 1.26 hours. You initially plan to take an SRS of size n = 165 and compute the average total sleep time.

Answer 1

Answer:

a) $\sigma_{\bar X}= \frac{1.26}{\sqrt{165}}= 0.0981$

b) We expect 68% of the values between : (6.6219, 6.8181)

We expect 95% of the values between : (6.5238, 6.9162)

We expect 99.7% of the values between : (6.4257, 7.0143)

c) $z = \frac{6.9-6.72}{0.0981}= 1.835$

And we can use the z normal table or excel and we got:

$P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the total sleep time of a population, and for this case we know the distribution for X is given by:

$X \sim N(6.72,1.26)$  

Where $\mu=6.72$ and $\sigma=1.26$

(a)  What is the standard deviation for the average time in hours? (Round your answer to four decimal places.)

For this case the sample mean have the following distirbution

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the deviation is given by:

$\sigma_{\bar X}= \frac{1.26}{\sqrt{165}}= 0.0981$

in minutes?= (Round your answer to three decimal places.)  

(b)  Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean. (Round your answer to four decimal places.)

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

We expect 68% of the values between : (6.6219, 6.8181)

We expect 95% of the values between : (6.5238, 6.9162)

We expect 99.7% of the values between : (6.4257, 7.0143)

(c) What is the probability that your average will be below 6.9 hours? (Round your answer to four decimal places.)

For this case we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And we can find the z score for 6.9 and we got:

$z = \frac{6.9-6.72}{0.0981}= 1.835$

And we can use the z normal table or excel and we got:

$P(Z$