Answer:
the standard deviation increases
Step-by-step explanation:
Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student
The sample means = x₁ + x₂ + . . . , x₂₃ / 23
[tex]= \frac{x_1 +x_2 +...x_2_3}{23}[/tex]
= 2.3hr
⇒[tex]\sum xi =2.3 \times 23 = 52.9hrs[/tex]
let x₁ , x₂, . . . , x₂₃ arranged in ascending order
Then x₂₃ was 10 and has been changed to 14
i.e x₂₃ increase to 4
Sample mean [tex]= \frac{x_1 +x_2 +...x_2_3}{23}[/tex]
[tex]\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47[/tex]
therefore, the new sample mean is 2.47
2) For the old data set
the median is [tex]x_1_2(th)[/tex] values
[tex][\frac{n +1}{2} ]^t^h value[/tex]
when we use the new data set only x₂₃ is changed to 14
i.e the rest all observation remain unchanged
Hence, sample median = [tex][{x_1_2]^t^h value[/tex] remain unchange
sample median = 2.5hrs
The Standard deviation of old data set is calculated
[tex]=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)[/tex]
The new sample standard sample deviation is calculated as
[tex]= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)[/tex]
Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.
Since,
(xi - 2.3)² ∑ (xi - 2.47)²
Therefore , the standard deviation increases
