Pls help im talkin 25 here

Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.

Georgia [21]

Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

$z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ 
z-score =2.326$

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.

5 0

3 years ago

An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 180180 engin

tiny-mole [99]

Answer:

$z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98$

$p_v =P(Z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=4.6$ represent the sample mean

$\sigma=0.9$ represent the population standard deviation

$n=180$ sample size

$\mu_o =4.8$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if thetrue mean is below the specifications, the system of hypothesis would be:

Null hypothesis: $\mu \geq 4.8$

Alternative hypothesis: $\mu < 4.8$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98$

P-value

Since is a one sided test the p value would be:

$p_v =P(Z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

4 0

3 years ago

A drawbridge rises at a rate of 3 2/3 feet per minute. The drawbridge rises to a height of 50 feet. How long does it take the dr

k0ka [10]

Answer: Third option.

Step-by-step explanation:

You know that the drawbridge rises at a rate of $3\ \frac{2}{3}$ feet per minute. This means that the drawbridge rises $3\ \frac{2}{3}$ in a minutes.

Convert from mixed fraction to decimal number:

$3\ \frac{2}{3}\ ft=(3+0.66)\ ft=3.6\ ft$

Since the drawbridge rises to a height of 50 feet, you need to apply this procedure in order to find the time the drawbridge takes to completely rise:

$time=\frac{(1\ min)(50\ ft)}{3.66\ ft}=13.6\ min$ or $13\ \frac{7}{11}\ min$