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weeeeeb [17]
3 years ago
12

Suppose on a certain test, fifty students earn a perfect score and fifty students earn a zero. determine the mean grade and stan

dard deviation among the one hundred students.
a. mean grade among the one hundred students is a0
b. standard deviation:
Mathematics
2 answers:
kodGreya [7K]3 years ago
5 0

Answer:

Mean=50.

standard deviation=50.

Step-by-step explanation:

Let the perfect score be 100.

The mean(average) of a data is given by: Mean=\frac{sum of data points}{number of data points}

Here the number of data points are 100. out of which 50 attains a value 100,and 50 attains value 0.

so, sum of data points=50×100+50×0=5000.

Mean=\frac{5000}{100}

Mean=50.

"Now the standard deviation of data points are calculated by firstly subtracting mean from every entry and then square the number and take it as new entry and calculate the mean of the new data entry and lastly taking the square root of this new mean".

Here if 50 is subtracted from each entry the new entry will have 50 entries as '50' and 50 entries as '-50'.

next on squaring we will have all the 100 entries as '2500'.

now the mean of these entries is: \frac{2500\times100}{100}

                                                         =2500

taking it's squareroot we have \sqrt{2500}=50

Hence, standard deviation=50.




yKpoI14uk [10]3 years ago
5 0

Answer:

mean is 50

standard deviation is 50

Step-by-step explanation:

as for finding the mean, if half get 100 and half get nothing, all you do is divide 100 by 2. that gets you 50.

as for standard deviation:

first you want to subtract the mean (fifty) from all your numbers.

100-50=50\\0-50=-50 (remember we still have fifty of each)

now square those numbers you have

50^{2} = 2500\\(-50)^{2}=2500

now's the time to remember you have fifty of each value. go ahead and multiply fify into each.

50*2500=125,000\\50*2500=125,000

now add em up

125,000+125,000= 250000

divide by 100

250000/100= 2500

now get the square root of that

\sqrt{2500}=50

and boom, you have your standard deviation.

hope this helped

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