HELP PLSSSS LOL!!!!!!!

For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 50

Volgvan

Answer:

$P(\bar X>530)=1-0.808=0.192$

Step-by-step explanation:

1) 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 scores, and for this case we know the distribution for X is given by:

$X \sim N(\mu=505,\sigma=170)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

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

On this case $\bar X \sim N(505,\frac{170}{\sqrt{35}})$

2) Calculate the probability

We want this probability:

$P(\bar X>530)=1-P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(\bar X >530)=1-P(Z$

$P(\bar X>530)=1-0.808=0.192$

3 0

3 years ago

HELP WILL GIVE HIGHEST RATING!!!

pychu [463]

The practical domain would be the number of rolls she can get which is between 0 and 10

6 0

3 years ago

Read 2 more answers

Help me pleaseee!! I’ll mark

sukhopar [10]

Answer:

esarespuesta ad 56

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

$0.25 per bottle water

sweet-ann [11.9K]

Answer:

24-pack

Step-by-step explanation:

you are paying less per bottle than the other 2 packs.

4 0

3 years ago

An accounting professor is notorious for being stingy in giving out good letter grades. In a large section of 140 students in th

Arlecino [84]

Answer:

The 95% confidence interval for the proportion of students that obtain a letter grade of B or better from this professor is (0.2056, 0.3544). The interpretation is that we are 95% sure that the true proportion of students who obtain a letter grade of B or better from this professor is between 0.2056 and 0.3544.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

140 students, so $n = 140$

B or better are grades of A or B.

5% earn As, 23% earn Bs, so $p = 0.05 + 0.23 = 0.28$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.96\sqrt{\frac{0.28*0.72}{140}} = 0.2056$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.96\sqrt{\frac{0.28*0.72}{140}} = 0.3544$

The 95% confidence interval for the proportion of students that obtain a letter grade of B or better from this professor is (0.2056, 0.3544). The interpretation is that we are 95% sure that the true proportion of students who obtain a letter grade of B or better from this professor is between 0.2056 and 0.3544.

7 0

3 years ago