Materials needed to answer the questions below:

$"""$ You can use one of two approaches:

$1)$ add suitable code below and then run this file

$2)$ run this file first then do the calculation in the

# LFS: ILO redundancy level $($thousands$)$: UK$,1995$ to $2022$

# Source dataset ID LMS

# Unit $000\text{'}$s

# LFS: ILO redundancy level $($thousands$)$: UK: Men

redundancy$\_$men$=[108,110,101,111,116,107,115,121,104,88,91,90,78,106,$

$156,100,85,92,74,62,65,68,61,50,60,125,56,46]$

# LFS: ILO redundancy level $($thousands$)$: UK: Women

redundancy$\_$women$=[67,57,54,59,62,63,62,64,54,53,51,48,50,59,79,55,63,$

$58,51,46,43,46,42,43,46,103,50,28]$

import math

def median$($alist$)$:

$"""$ Calculates the median of a list of numbers.

The list must not be empty.

$"""$

number$\_$of$\_$values $=$ len$($alist$)$

sorted$\_$list $=$ sorted$($alist$)$

# Two cases, depending on whether the number of values is odd or even.

quotient $=$ number$\_$of$\_$values $//2$

remainder $=$ number$\_$of$\_$values $\%2$

if $($remainder $==1)$:

result $=$ sorted$\_$list$[$quotient$]$

else:

result $=($sorted$\_$list$[$quotient $-1]+$ sorted$\_$list$[$quotient$])/2$

return result

def test$\_$median$()$:

assert median$([2])==2$

assert median$([4,3])==3\mathrm{.}5$

assert median$([3,1,8,4,7,6,4,2,5,9])==4\mathrm{.}5$

assert median$([7,2,6,2,5,3,1,0,8,6,6,4,9])==5$

# Unit test

test$\_$median$()$

def mean$($list$)$:

$"""$Return mean of list"""

sum $=0$

count $=0$

for item in list:

sum $=$ sum $+$ item

count $=$ count $+1$

return sum $/$ count

def test$\_$mean$()$:

list $=[1,2,3,4,5]$

assert$($mean$($list$)==3)$

# Unit test

test$\_$mean$()$

def corr$\_$coef$($list$\_$x$,$ list$\_$y$)$:

$"""$ Return correlation between values in list$\_$x and list$\_$y$.$

Lists must be of equal length.

$"""$

x$\_$bar $=$ mean$($list$\_$x$)$

y$\_$bar $=$ mean$($list$\_$y$)$

sxy $=0$

sxx $=0$

syy $=0$

for index in range$($len$($list$\_$x$))$:

x $=$ list$\_$x$[$index$]$

y $=$ list$\_$y$[$index$]$

sxy $=$ sxy $+($x $-$ x$\_$bar$)*($y $-$ y$\_$bar$)$

sxx $=$ sxx $+($x $-$ x$\_$bar$)*($x $-$ x$\_$bar$)$

syy $=$ syy $+($y $-$ y$\_$bar$)*($y $-$ y$\_$bar$)$

return sxy $/$ math.sqrt$($sxx $*$ syy$)$

def test$\_$corr$\_$coef$()$:

# Data from M$140$ Unit $9$ Example $5$

list$1=[78\mathrm{.}9,75\mathrm{.}8,77\mathrm{.}3,74\mathrm{.}2,78\mathrm{.}1,72\mathrm{.}8,77\mathrm{.}6,77\mathrm{.}9]$

list$2=[56\mathrm{.}7,53\mathrm{.}1,56\mathrm{.}1,55\mathrm{.}9,54\mathrm{.}1,48\mathrm{.}6,59\mathrm{.}4,54\mathrm{.}0]$

assert round$($corr$\_$coef$($list$1,$ list$2),2)==0\mathrm{.}64$

# Unit test

test$\_$corr$\_$coef$()$

Questions:

a$)$

i$.$ Why might it be important to consider median redundancy levels in addition to mean values? How might outliers affect the interpretation of mean values compared to median values? Use information from attached picture.

ii$.$

Use this function provided above to find the median of redundancy$\_$men.

In your Solution give the median, also provide the Python code you used for calling the median$()$ function and explain how you executed it$.$

iii. The median redundancy level for women is $54,000.$ Identify two factors that may contribute to the difference in median redundancy levels between men and women in the United Kingdom. Use provided materials from above

b$)$

i$.$ Use function provided above to calculate the correlation coefficient between redundancy$\_$men and redundancy$\_$women.

In your Solution, provide the resulting figure rounded $($manually or using Python$)$ to three decimal places. Also provide the Python code you used for calling the corr$\_$coef$()$ function and explain how you executed it$.$

ii$.$Briefly discuss whether any effect you have found is likely to be causal.