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EleoNora [17]
3 years ago
15

A study uses a sample of 200 college students to learn about the number of hours they studied for their final exams. The mean nu

mber of hours was 136 and the standard deviation was 14 hours. If this data falls under a normal distribution or bell curve, which of the following statements is NOT true?
The number of students that studied 122 or fewer hours was about 32.

The number of students that studied 150 or fewer hours was about 168.

The majority of students studied between 122 and 150 hours.

No one studied less than 108 hours.
Mathematics
2 answers:
ZanzabumX [31]3 years ago
6 0
\displaystyle\mathbb P(X\le122)=\mathbb P\left(\frac{X-136}{14}\le\frac{122-136}{14}\right)=\mathbb P(Z\le-1)\approx0.1587
which means 0.1587\times200\approx32 studied 122 or fewer hours.

\displaystyle\mathbb P(X\le150)=\mathbb P\left(\frac{X-136}{14}\le\frac{150-136}{14}\right)=\mathbb P(Z\le1)\approx0.8413
which means 0.8413\times200\approx168 studied 150 or fewer hours.

So, neither of the first two are false.

Depending on your definition of "majority", you could also eliminate the third option. From the above calculations, you know that students that studied between 122 and 150 hours fall within 1 standard deviation of the mean, which corresponds to approximately 65% of the student population. "Majority" is often taken to mean "more than 50%", so this is also true.

This leaves us with the last option. For fun, let's compute that probability.
\displaystyle\mathbb P(X
which means 0.0228\times200\approx5 students studied less than 108 hours.
Dovator [93]3 years ago
4 0

Answer:

No one Studied less than 108 hours

Step-by-step explanation:

This is the answer because there were about 5 people that did study less than 108 hours! Hope this makes it easier to read

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A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.
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Answer:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean  is also normal and given by:

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

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

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".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Solution to the problem

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

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean \bar X is also normal and given by:

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

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

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3 years ago
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