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Ber [7]
2 years ago
13

A data set has a maximum of 84, a minimum of 34, a median of 52.5, an upper quartile of 64, and a lower quartile of 44. Which st

atement is true of the box plot of this data set? The box will go from 44 to 64.A line dividing the box will be at 54.The left whisker will go from 34 to 52.5.The right whisker will go from 44 to 84.
Mathematics
1 answer:
Akimi4 [234]2 years ago
4 0

Answer:

A box and whiskers plot has a box whose endpoints are the interquartile range (25th percentile and 75th percentile), has a dividing line at the median, aka the 50th percentile, and whiskers that extend out to the minimum and the maximum of the data.

Let's sort the info for ease of reference:

min 34, lower quartile 44, median 52.5, upper quartile 64, max 84.

OK, let's go through the statements.

A. The box will go from 44 to 64.

The box goes from the lower quartile to the upper quartile, TRUE

B. A line dividing the box will be at 54.

There's no 54 mentioned at all, FALSE

C. The left whisker will go from 34 to 52.5.

The left whisker goes from the minimum to the lower quartile, 34 to 44 here, FALSE.

D. The right whisker will go from 44 to 84.

The right whisker goes from the upper quartile to the max, 64 to 84 here.  FALSE

If this could help, please mark me brainliest! Have an awesome day too!

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the parent function of the function g(x)=(x-h)^2+k is f(x)=x^2. the vertex of the function g(x) is located at (9,8). what are th
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Answer:

<h2>h = 9, k = 8</h2>

Step-by-step explanation:

\text{The vertex form of an an equation of a quadratic function:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\\\\text{We have}\ g(x)=(x-h)^2+k,\ \text{and the vertex in}\ (9,\ 8).\\\\\text{Therefore}\ h=9\ \text{and}\ k=8.\ \text{substitute:}\\\\g(x)=(x-9)^2+8

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At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul
vampirchik [111]

Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b) The 95% confidence interval would be given by (910.05;959.95)    

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=3278.222

The sample deviation calculated s=97.054

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

So on this case the 95% confidence interval would be given by (910.05;959.95)    

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

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

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

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Answer:

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