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2 years ago

What is the interquartile range (IQR) of the data set represented by this box

Mathematics

1 answer:

larisa [96]2 years ago

8 0

Answer:

36 :)

Step-by-step explanation:

First put the numbers in order from least to greatest.

0, 10, 20, 30, 32, 40, 42, 50, 56, 59, 60.

The median is 40 (middle)

Find the median of the two quartiles. 56 and 20.

Now subtract them. your answer is 36.

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Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi

boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

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".

Description in words of the parameter p

$p$ represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that have experienced bullying

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{63}{200}=0.315$ represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261$

$0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369$

And the 90% confidence interval would be given (0.261;0.369).

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).

lower limit =0.261

upper limit =0.369

5 0

3 years ago

Question 14 of 20 :

LenKa [72]

Where is the graph for

5 0

3 years ago

Someone explain how to do this please!! dont need just an answer but explain as well

Leya [2.2K]

6.5 goes in the first box and 5.5 goes in the second box.

Explanation for first box: 13+2 is 15 and half of 15 is 7.5 so you need to take 13 and subtract 7.5. Once you do that, you end up with 6.5

Explanation for second box: (The negative 7 doesn't matter because you want the absolute value) 18+7 is 25 and half of 25 is 12.5. 18-12.5 is 5.5 so the answer would be 5.5

8 0

2 years ago