Please help to solve number 3 only

Please help to solve number 3 only
If the age of a sample of seven employees of a fast-food outlet (X) are as follows
19, 19, 65, 20, 21, 18, 20
1) Compute the best central measure and the best absolute dispersion measure.
2) If the oldest employee is retired, compute the best central measures and the best absolute dispersion measure.
3) If the employees in this outlet has a mean of age = 26 with variance between their ages as =16. Find the mean and the standard deviation of the age of employees of a fast-food in another outlet (Y), if the relation between them is as follows
Y = - 0.5 X + 35

Thanks in advance
i think Should i replace each value of Y with X_i i.e Y= 25.5, 25.5, 2.5, 25, 24.5, 26,25

AhmedFathy said:

Please help to solve number 3 only
If the age of a sample of seven employees of a fast-food outlet (X) are as follows
19, 19, 65, 20, 21, 18, 20
1) Compute the best central measure and the best absolute dispersion measure.
2) If the oldest employee is retired, compute the best central measures and the best absolute dispersion measure.
3) If the employees in this outlet has a mean of age = 26 with variance between their ages as =16. Find the mean and the standard deviation of the age of employees of a fast-food in another outlet (Y), if the relation between them is as follows
Y = - 0.5 X + 35

Thanks in advance
i think Should i replace each value of Y with X_i i.e Y= 25.5, 25.5, 2.5, 25, 24.5, 26,25

Click to expand...

Yes, you do replace the initial values with the Y's you have. However, have you yet been told that, under a linear transformation, the mean is transformed the same way and the standard deviation is multiplied by the magnitude of the slope? So the new mean is -0.5 * 26 + 35 = 22 and the new standard deviation is 0.5 * 16 = 8 without having to go through the calculations. You might try to prove this as an aid to keeping it in mind.

Ishuda said:

Yes, you do replace the initial values with the Y's you have. However, have you yet been told that, under a linear transformation, the mean is transformed the same way and the standard deviation is multiplied by the magnitude of the slope? So the new mean is -0.5 * 26 + 35 = 22 and the new standard deviation is 0.5 * 16 = 8 without having to go through the calculations. You might try to prove this as an aid to keeping it in mind.

Click to expand...

Thanks for your quick response, this is the third lecture in statistics in MBA, accroding to my information until now..
we cant calculate the standard deviation for Y because it contains low extreme value so its skewed data.. any way even after calculation the standard deviation = 8.61
please reply to check my understand

AhmedFathy said:

Thanks for your quick response, this is the third lecture in statistics in MBA, accroding to my information until now..
we cant calculate the standard deviation for Y because it contains low extreme value so its skewed data.. any way even after calculation the standard deviation = 8.61
please reply to check my understand

Click to expand...

Sorry, but the answer is yea and no, depending on just what you mean.

Forgetting for a moment, that the 2.5 year old working in a fast-food outlet is extremely strange to say the least, the standard deviation of that complete set is defined (and is the 0.5 times the standard deviation of the other set). So, in one sense, you are wrong when you say you can't compute the standard deviation of that set. It's just that it may not represent the real word very well.

However, in the real world, any data set should be looked at to see if there are errors in the data (numbers transcribed incorrectly, data from a different population, etc). In that sense, one would throw out the data point for the 2.5 year old kid working in a fast food outlet (surely it is a mistake) and compute the mean and standard deviation of a new data set consisting of all the old values minus the ones you threw away. In that case the new mean and standard deviation may not be easily related to the old values. So, in another sense, you are correct when you say 'the data, as it stands, doesn't make sense and you can't compute the mean and standard deviation'.

For a class, which approach is used to determine the mean and standard deviation is determined by the class and instructor. However, if you use the second approach (eliminating unsound data), it should be explained in the results why you eliminated the data.