The time spent by 60 students on social media in a week is recorded. The cumulative frequency curve below shows the distribution of the data collected.

1. Identify the median point that must remain unchanged.

Since there are $60$ students, the median is the value at cumulative frequency
$\frac{60}{2}=30.$
The new curve must pass through the same point on the horizontal line $\text{CF}=30$ as the original curve.

2. Ensure the interquartile range is larger than the original distribution.

For $60$ students,
$Q_1\text{ is read at CF }=15, \qquad Q_3\text{ is read at CF }=45.$
A larger interquartile range means
$Q_3-Q_1\text{ is larger}.$
So the new curve should meet $\text{CF}=15$ at a smaller time than the original curve and meet $\text{CF}=45$ at a larger time than the original curve.

3. Sketch a valid cumulative frequency curve.

Draw a smooth increasing curve from near cumulative frequency $0$ to cumulative frequency $60$. The curve should:

• pass through the same median point as the original curve at $\text{CF}=30$,
• be further to the left of the original curve at $\text{CF}=15$,
• be further to the right of the original curve at $\text{CF}=45$,
• remain non-decreasing throughout.

1. Identify the number of girls who spent at most 60 minutes and form the probability.

Girls who spent at most $60$ minutes are in the classes $20 < x \leq 40$ and $40 < x \leq 60$.

Number of such girls $= 8 + 30 = 38$.

Total number of students $= 240$.

$P(\text{girl and at most }60\text{ min})=\frac{38}{240}=\frac{19}{120}$

1. Find the number of students who spent more than 80 minutes and form the probability.

Students who spent more than $80$ minutes are in the class $80 < x \leq 100$.

Number of such students $= 5 + 40 = 45$.

Total number of students $= 240$.

$P(\text{spent more than }80\text{ minutes})=\frac{45}{240}=\frac{3}{16}$

1. Find the number of students who spent less than or equal to 40 minutes.

From the table, students who spent $x \leq 40$ minutes:

$15+8=23$

So the number who spent more than $40$ minutes is

$240-23=217.$

2. Use the complement probability for at least one student.

$P(\text{at least one spent } \leq 40) = 1 - P(\text{both spent } >40)$

Since the students are selected without replacement,

$P(\text{both spent } >40)=\frac{217}{240}\times\frac{216}{239}.$

Hence

$P(\text{at least one})=1-\frac{217}{240}\times\frac{216}{239}$

$=1-\frac{46872}{57360}$

$=\frac{10488}{57360}=\frac{437}{2390}.$