Question

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours and a median of 2.5 hours. When they review their data, they notice that the maximum value was recorded as 10 hours and it should have been 14. they fix this error. When compare to origional dataset, did the IQR:______________.1. Increase2. Decrease3. Stayed the same4. There is no way to tell without actually seeing the data

Answer 1

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

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒$\sum xi =2.3 \times 23 = 52.9hrs$

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  $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

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