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2 years ago

What is the interquartille range of the given data set

Mathematics

1 answer:

IrinaK [193]2 years ago

3 0

Answer: B. 9

Step-by-step explanation:

First, find the median of the data set. This set has an even number of points, so find the average between the two middle points: 18 and 19. 18+19 = 37. 37/2 = 18.5. The median is 18.5.

Now, to find the lower quartile, find the median of the lower half of the data set {11, 12, 14, 15, 18}. The number in the middle is 14. The lower quartile is 14.

To find the upper quartile, find the median of the upper half of the data set {19, 21, 23, 25, 55}. The number in the middle is 23. The upper quartile is 23.

To find the interquartile range, subtract the lower quartile from the upper quartile. 23-14 = 9. The interquartile range is 9.

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5m-8=2m+2 what does m equal

vesna_86 [32]

Answer:

m = 10/3

(as a decimal: m = 3.333...)

Step-by-step explanation:

5m - 8 = 2m + 2

- 2m - 2m (make all "m" on one side to isolate variable)

3m - 8 = 2

+ 8 + 8 (get m entirely on 1 side)

3m = 10

÷3 ÷3 (divide by 3 to get 1m)

m = 10/3

(as a decimal: m = 3.333...)

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Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teache

Tju [1.3M]

Answer:

The 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

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".

$p_A$ represent the real population proportion for elementary school

$\hat p_A =\frac{226}{397}=0.569$ represent the estimated proportion for elementary school

$n_A=397$ is the sample size required for Brand A

$p_B$ represent the real population proportion for high school teachers

$\hat p_B =\frac{129}{268}=0.481$ represent the estimated proportion for high school teachers

$n_B=268$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.569-0.481) - 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.0109$

$(0.569-0.481) + 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.1651$

And the 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

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3 years ago