4. Find f(2) if f(x) = 2x + 1.

For this case we want to conduct a test in order to see if the proportion of Clemson students who eat breakfast is different from 0.44 (alternative hypothesis). And the complement would represent the null hypothesis, and the system of hypothesis for this case are:

Null Hypothesis: $p =0.44$

Alternative hypothesis: $p \neq 0.44$

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

Solution to the problem

For this case we want to conduct a test in order to see if the proportion of Clemson students who eat breakfast is different from 0.44 (alternative hypothesis). And the complement would represent the null hypothesis, and the system of hypothesis for this case are:

Null Hypothesis: $p =0.44$

Alternative hypothesis: $p \neq 0.44$

8 0

3 years ago

The physical plant at the main campus of a large state university receives daily requests to replace fluorescent light bulbs. Th

ankoles [38]

We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.

First of all, we will find z-score corresponding to 38 and 56.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{38-38}{6}=\frac{0}{6}=0$

Now we will find z-score corresponding to 56.

$z=\frac{56-38}{6}=\frac{18}{6}=3$

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is $-3\sigma\text{ to }3\sigma$ .

We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.

We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

$\frac{99.7\%}{2}=49.85\%$

Therefore, approximately $49.85\%$ of lightbulb replacement requests numbering between 38 and 56.

6 0

3 years ago