The standard deviation of the data sample give the degree of spread of the data. Hence, the standard deviation of the data is 0.08
The sample standard deviation is defined as :
[tex]\sqrt {\frac{(x - \mean)^{2} }{n-1}} [/tex]The mean :
[tex] \mean = \frac{ x_{1} + x_{2} +... + x_{n}}{n} [/tex][tex] \mean = \frac{0.4 + 0.24 + 0.22 + 0.26 + 0.34}{5} = 0.292[/tex]
The Variance :
[tex]\sqrt {\frac{(0.4 - 0.292)^{2} + (0.24 - 0.292)^{2} + (0.22 - 0.292)^{2} + (0.26 - 0.292)^{2} + (0.34 - 0.292)^{2}}{5 - 1}} = 0.00572[/tex]
The standard deviation :
[tex]\sqrt {\frac{(x - \mean)^{2} }{n-1}} [/tex]
[tex]\sqrt {0.00572} = 0.0756[/tex]
Therefore, the standard deviation of the data sample is 0.08.
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Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the 20th, 25th,65th , and 75th percentiles
Answer:
Percentile of 20th = 20
Percentile of 25th = 22.5
Percentile of 65th = 28
Percentile of 75th = 29
Explanation:
Sort the numbers in ascending order:
15,20,25,25,27,28,30,34
i = P/100 * N ( i index is the position of the number. Index 1 = 15, index 2 = 20, and so on)
Index of the 20th percentile:
20/100 * 8 = 1.6
Round 1.6 to the next integer = 2
The second number I=20
The 20th percentile is 20
Index of the 25th percentile:
25/100 * 8 = 2
Because 2 is an integer the percentile of 25th is the average of I=20 and I=25
(20+25)/2=22.5
The 25th percentile is 22.5
Index of the 65th percentile
i = 65/100 * 8 = 26.5 = 5.2
Round to the next integer = 6
The 65th percentile is 28
Index of the 75th percentile
I = 75/100 * 8 = 6
Since 6 is an integer (28+30)/2 = 29 percentile
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Which describes a peer review? Reading your own writing and making changes. Have your friends or fellow students read your writing and suggest changes. Have your learning coach read your writing to make changes. Reading the story to your pet.
Answer:
have your friends read and make suggestions
Explanation:
peer review means to have your classmates read your work and help you make changes.
Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question.
Listed below are the jersey numbers of 11 players randomly selected from the roster of a championship sports team. What do the results tell us?
24 26 89 16 38 69 93 3 1 80 49
The statistical values such as mean, Median and mode gives measure of centre of a data, while the midrange gives information about the spread. Hence, the statistical measures are :
Mean = 44.36Median = 38Mode = 1, 3, 16, 24, 26, 38, 49, 69, 80, 89, 93Midrange = 46Given the data :
1, 3, 16, 24, 26, 38, 49, 69, 80, 89, 93Mean = [Σx ÷ n]
Mean = (24+26+89+16+38+69+93+3+1+80+49) / 11
Mean = (488 ÷ 11) = 44.36
The median :
0.5(n + 1)th termMedian = 0.5(11 + 1) = 6th term
Median = 38
The Mode :
The most frequently occurring value in a distribution ; since each the value occur just one, then there is no single mode value.
The midrange :
(maximum - minimum) / 2Midrange = (93 - 1) / 2
Midrange = 92/2 = 46
Therefore, the midrange is the value of the range divided by 2.
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FILIPINO
basahin ang teksto sa kahon dugtungan ang paungan salit bawat talata iyong pagsasalaysay
Answer:
plss pa sagot po
Explanation:
thanks sa points
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