a)
The mean waiting time per child can be calculated using the formula:
![\mu=(1)/(N)\sum_{k\mathop{=}1}^na_k\cdot n_k](https://img.qammunity.org/2023/formulas/mathematics/college/sd9lpc4hsvsf1y7q7t33xlv2y7gy2thzqs.png)
Where a_k is the time, n_k is the number of children for time k, and N is the total number of children. Using the data from the problem:
![\begin{gathered} \mu=(10\cdot40+20\cdot50+30\cdot10+40\cdot70+50\cdot0+60\cdot30)/(40+50+10+70+0+30) \\ \\ \mu=(400+1000+300+2800+0+1800)/(200)=(6300)/(200) \\ \\ \therefore\mu=31.5\text{ minutes} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d97dg91takg5rebk1kzwja716mlm62xqae.png)
b)
For the median, we calculate the tentative position of the median using the equation:
![i=(N)/(2)=(100)/(2)=100](https://img.qammunity.org/2023/formulas/mathematics/college/3afg6ouyl2b854vajhrg2iqiju1z1pcl1a.png)
Since this is an integer number, the real median is the semi-sum of the elements at positions i and (i+1). From the table, these elements are 30 and 40, so the median is:
![\begin{gathered} Q_2=(30+40)/(2)=(70)/(2) \\ \\ \therefore Q_2=35\text{ minutes} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/52sujmpbvlkoxqw8ub9mwgmh1vs84rhfi6.png)