Answer: a) (0.755, 0.925)
Step-by-step explanation:
Let p be the population proportion of 8th graders are involved with some type of after school activity.
As per given , we have
n= 100
sample proportion: ![\hat{p}=0.84](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%3D0.84)
Significance level : ![\alpha= 1-0.98=0.02](https://tex.z-dn.net/?f=%5Calpha%3D%201-0.98%3D0.02)
Critical z-value :
(using z-value table)
Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-
![\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%5Cpm%20z_%7B%5Calpha%2F2%7D%5Csqrt%7B%5Cdfrac%7B%5Chat%7Bp%7D%281-%5Chat%7Bp%7D%29%7D%7Bn%7D%7D)
i.e. ![0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}](https://tex.z-dn.net/?f=0.84%5Cpm%20%282.33%29%5Csqrt%7B%5Cdfrac%7B0.84%281-0.84%29%7D%7B100%7D%7D)
i.e. ![\approx0.84\pm 0.085](https://tex.z-dn.net/?f=%5Capprox0.84%5Cpm%200.085)
i.e. ![(0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)](https://tex.z-dn.net/?f=%280.84-%200.085%2C%5C%200.84%2B%200.085%29%3D%280.755%2C%5C%200.925%29)
Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)
12k - 2k +16 you add the 2 and 10 on the first half and the 3 and 13 on the other half
You subtract and find x=10
I can't understand what you wanted with the second question (or if that is even a question), but the answer to the first one is 42.