Given a sample: 6, 8, 13, 10, 7, x and knowing that the median is equal to 9. Find the possible values of x

Question

Given a sample: 6, 8, 13, 10, 7, x and knowing that the median is equal to 9. Find the possible values of x.

The answer is x≥10

Answer 1

Review the definitions of MEAN, MODE, and MEDIAN.

Median, in statistics, is the middle value of the given list of data, when arranged in an ascending or descending order.

If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values

So, sort the numbers to get: 6, 7, 8, 10, 13

There will be six (an even number), so a mean of 9 is the mean of 8 and 10 [i.e., average of those numbers That makes the list look like:

6, 7, 8, 10, x, 13

or 6, 7, 8, 10, 13, x

x could be another 10 or it could be any number greater than 10 (note: sorted list).

ANSWER: x ≥ 10

Answer 2

6,7,8,10, 13

have median = (8+10)/2 = 18/2 = 9, x> 10 is above the median

mean = (6+7+8+10+13)/5 = 44/5 = 6.8

the distribution is assymetric, skewed to the right, due to large nearly outlying data

variance = sum of squared differences from the mean divided by n

= 70.64/5 = 14.128

standard deviation = sqr14.128 = about 3.159

8.8+3.2 =12

10 is less than one sd from the mean

and 13 is less than 2 sd from the mean but close.

if x is greater than or equal to 10, x is also less than or equal to 1 standard deviation from the mean

if 10<x<13, x is < 1 sd from the mean

if x > 13, x is > 1 sd

if x > 21.2 x > 2 sd

if x > 25.4, x > 3 sd and is a likely outlier, unreliable or possibly impossible

Except if x is a 6th number > or =10

recalculate the mean and variance

6,7,8,10,10, 13 have a mean of 54/6 = 9

variance then = 32/6 = 5 1/3

sd = sqr5 1/3 =2.309 =about 2.3

then 9+2.3 = 11.3= 1 sd from the mean

11.3+2.3= 13.6 = 2 sd from the mean

then 13 may be an outlier, almost = 2 sd from the mean

if x > 10

but x < 22

and x is not an outlier

IF x= 22

mean = 11

sd = 5.4

11+10.8 = 21.8. 22 itself may be an outlier as x is more than 2 sd from the mean

x<22 to be a reliable data point

It's unclear what the problem is asking, but

IF the problem is asking what value the median is when x is 6th data point

then group them in order, 6,7,8, x, 10, 13, and then x = between 8 and 10

or if x > 10, the median also is 9, no matter how large x is

6,7,8,10,10,13 has median = 9

6,7, 8, 10, x, 13 with x between 10 and 13 also has median = 9

6,7,8,10,13,x also has median = 9 as the average of 8 and 10 is still the median =9

for the median to remain 9, x > 10. x is either 10 or any number larger than 10, then the median remain = 9

Sometimes the problem is figuring out what the problem is asking.