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4 years ago

Between which two thousands is the number 36,219

2 answers:

4 0

32,000 and 42,000. I know this because the question states "Two Thousands." 32,000 and 42,000 are both two thousands because 32 and 42 both end with 2's because of the "Two" in "Two Thousand."

il63 [147K]4 years ago

4 0

Between 35,200 because if you can see. That if. You look the number said. 36,219. It would be 35,200

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The sample mean is $\bar{x}=\frac{1183}{18}\approx 65.722$ .

The sample standard deviation is $s=\frac{\sqrt{1666105}}{105}\approx 12.293$ .

The lower quartile is 59.

The upper quartile is 75.

The median is $\frac{135}{2}=67.5$ .

Step-by-step explanation:

We have the following data:

83, 45, 61, 40, 83, 67, 45, 66, 70, 69, 80, 58, 68, 60, 67, 72, 73, 70, 57, 63, 70, 78, 52, 67, 53, 67, 75, 61, 70, 81, 76, 79, 75, 76, 58, 31.

(a)

The sample mean of data is given by the formula

$\large{{\overline{{{x}}}}=\frac{{1}}{{n}}{\sum_{{{i}={1}}}^{{n}}}{x}_{{i}}}$ ,

where n is the number of values and $\large{{x}_{{i}},{i}={\overline{{{1}..{n}}}}}$ are the values themselves.

Since we have 36 points, then n = 36.

The sum of the points is $\sum_{i=1}^{n} x_i=2366$ .

Therefore, the sample mean is $\bar{x}=\frac{1}{36}\cdot2366=\frac{1183}{18}$ .

The sample standard deviation of data is given by the formula

$\large{{s}=\sqrt{{\frac{{1}}{{{n}-{1}}}{\sum_{{{i}={1}}}^{{n}}}{{\left({x}_{{i}}-{\overline{{{x}}}}\right)}}^{{2}}}}}$ ,

where n is the number of values, $\large{{x}_{{i}},{i}={\overline{{{1}..{n}}}}}$ are the values themselves, and ${\overline{{{x}}}$ is the mean of the values.

The mean of the data is $\bar{x}=\frac{1183}{18}$ .

n = 36.

Sum of ${{\left({x}_{{i}}-{\overline{{{x}}}}\right)}}^{{2}}$ is $\sum_{i=1}^{n} (x_i-\bar{x})^2=\frac{47603}{9}$ .

Now,

$\frac{1}{n-1} \sum_{i=1}^{n} (x_i-\bar{x})^2=\frac{1}{36-1}\cdot\frac{47603}{9}=\frac{1}{35}\cdot\frac{47603}{9}=\frac{47603}{315}$

Finally,

$s=\sqrt{\frac{47603}{315}}=\frac{\sqrt{1666105}}{105}\approx 12.2931133127713$

(b)

The $p^{th}$ percentile is a value such that at least $p$ percent of the observations is less than or equal to this value and at least ( $100-p$ ) percent of the observations is greater than or equal to this value.

The first step is to sort the values.

The sorted values are 31, 40, 45, 45, 52, 53, 57, 58, 58, 60, 61, 61, 63, 66, 67, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 80, 81, 83, 83.

The lower quartile or 25th percentile is the median of the lower half of the data set.

Now, calculate the index $i=\frac{p}{100}\cdot n=\frac{25}{100}\cdot 36=9$

Since the index $i$ is an integer, the 25th percentile is the average of the values at the positions $i$ and $i+1$ .

The value at the position $i=9$ is 58 and at the position $i+1=10$ is 60.

Their average is: $\frac{58+60}{2}=59$ .

The lower quartile is 59.

The upper quartile or 75th percentile is the median of the upper half of the data set.

Now, calculate the index $i=\frac{p}{100}\cdot n=\frac{75}{100}\cdot 36=27$

Since the index $i$ is an integer, the 75th percentile is the average of the values at the positions $i$ and $i+1$ .

The value at the position $i=27$ is 75 and at the position $i+1=28$ is 75.

Their average is: $\frac{75+75}{2}=75$ .

The upper quartile is 75.

(c) The median is the middle value of the data set. The median value depends on the number of values. If the number of values is odd, then the median is the "central" value among the sorted values. If the number of values is even, then the median is the average of the two "central values".

The first step is to sort the values.

We have 36 values, so their number is even.

Since the number is even, the median is the average of the "central values":

31, 40, 45, 45, 52, 53, 57, 58, 58, 60, 61, 61, 63, 66, 67, 67, 67, 67, 68, 69, 70, 70, 70, 70, 72, 73, 75, 75, 76, 76, 78, 79, 80, 81, 83, 83.

Calculate the median: $m=\frac{67+68}{2}=\frac{135}{2}$ .