Mathematics, 15.07.2020 01:01 haven765

suppose a person in the club with 91 members transfers to the club with 71 memebers. A student claims that the measures of center and the measures of spread will change. Correct the students error

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Answer from: oomale

It is an incorrect claim from the student that the measures of central tendency will change.

It is a correct claim from the student that the measures of spread will change but only range and standard deviation.

Step-by-step explanation:

The complete question is: Suppose a person in the club with 91 members transfers to the club with 71 members. A student claims that the measures of centers and the measures of spread will all change. Correct the student’s error.

Data set was: 80, 74, 77, 71, 75, and 91.

Firstly, we have to represent the original data set in ascending order;

Original data set = 71, 74, 75, 77, 80, and 91.

Now, it is stated that a person in the club with 91 members transfers to the club with 71 members, so the new data set is;

New data set = 72, 74, 75, 77, 80, and 90.

Now, first taking into account the measure of central tendency that is; Mean and Median.

Mean of both the data will remain the same because there is no change in the sum of all values in both the data set, i.e;

Mean of original data = $\frac{71+74+75+77+80+91}{6}$

= $\frac{468}{6}$ = 78

Mean of new data = $\frac{72+74+75+77+80+90}{6}$

= $\frac{468}{6}$ = 78

Now, the median of both the data set will also remain the same because there is a change in the first and last term of the data, so there will be no effect on the middle value in both the data.

Hence, it is an incorrect claim from the student that the measures of central tendency will change.

Now, taking into account the measure of spread that is; Range, Inter-quartile range, and Standard deviation.

Range is given by = Highest value - Lowest value

So, the range of original data = 91 - 71 = 20

and the range of new data = 90 - 72 = 18

This means that the range of data has been changed.

Now, the inter-quartile range = Second last term - Second term of data

So, the inter-quartile range of original data = 80 - 74 = 6

and the inter-quartile range of new data = 80 - 74 = 6

This means that the inter-quartile range of data has not been changed.

Now, the standard deviation = $\sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} }$

So, the standard deviation of the original data = $\sqrt{\frac{(71-78)^{2} +.......+ (91-78)^{2} }{6-1} }$

= 7.043

The standard deviation of the new data = $\sqrt{\frac{(72-78)^{2} +.......+ (90-78)^{2} }{6-1} }$

= 6.481

This means that the standard deviation has been decreased.

Answer from: Quest

about 5.6 days

step-by-step explanation:

we can fill the given information into the decay equation and solve for the value of t at which half the sample remains.

$\dfrac{11}{2} =11e^{-0.1247t} \qquad\text{decay equation with values filled in}\\\\\dfrac{1}{2}=e^{-0.1247t} \qquad\text{divide by 11}\\\\-\ln{(2)}=-0.1247t \qquad\text{take the natural log}\\\\\dfrac{\ln{(2)}}{0.1247}=t\approx 5.559\approx 5.6\qquad\text{divide by -0.1247}$