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777dan777 [17]
1 year ago
8

The scores on a statistics test had a mean of 81 and a standard deviation of 9. One student was absent on the test day, and his

score wasn’t included in the calculation. If his score of 84 was added to the distribution of scores, what would happen to the mean and standard deviation?.
Mathematics
1 answer:
Rama09 [41]1 year ago
8 0

Using the definitions of the mean and of the standard deviation of a data-set, we have that:

  • The mean would increase.
  • The standard deviation would remain constant.

<h3>What are the mean and the standard deviation of a data-set?</h3>

  • The mean of a data-set is given by the <u>sum of all values in the data-set, divided by the number of values</u>.
  • The standard deviation of a data-set is given by the <u>square root of the sum of the differences squared between each observation and the mean, divided by the number of values</u>.

When the measure of 84 is added to the data-set, we have that:

  • A measure greater than the mean is added, hence the mean would increase.
  • The difference squared between 84 and the mean of 81 is of 9, which is the same as the standard deviation, which would remain constant.

More can be learned about the mean and the standard deviation of a data-set at brainly.com/question/26941429

#SPJ1

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• let numbers be x, y and z

{ \tt{z = 8y - 9 -  -  - (eqn \: 1)}} \\  \\ { \tt{10x = 8y - 7 -  -  - (eqn \: 2)}} \\  \\ { \tt{x + y + z = 8 -  -  - (eqn \: 3)}}

• from eqn 2, make x the subject:

{ \tt{x =  \frac{8y - 7}{10} }} \\

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{ \tt{ \frac{8y - 7}{10}  + y + 8y - 9 = 8}} \\  \\ { \tt{8y - 7 + 10y + 80y - 90 = 80}} \\  \\ { \tt{98y = 177}} \\  \\ { \boxed{ \tt{ \: y = 1.8}}}

• find z

{ \tt{z = 8y - 9}} \\  \\ { \tt{z = 8(1.8) - 9}} \\  \\ { \tt{z = 14.4 - 9}} \\  \\ { \boxed{ \tt{ \: z = 5.4 \: }}}

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Rounding to nearest value:

{ \boxed{ \rm{x = 1}}} \\ { \boxed{ \rm{y =2 }}} \\ { \boxed{  \rm{z = 5}}}

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