Answer:
.
Explanation:
Given : Sample size : n= 64 , the sample is a large sample (n>30), so we can apply z-test.
Sample mean =
![\overline{x}=200](https://img.qammunity.org/2020/formulas/mathematics/college/c5sf87bd4jjm1k9qdpip0i5ka1iqp59asj.png)
Standard deviation :
![\sigma=48](https://img.qammunity.org/2020/formulas/mathematics/college/bur03snz2wz71n2omm943eo05uajlgj3c5.png)
Level of confidence:
![1-\alpha=0.95](https://img.qammunity.org/2020/formulas/mathematics/college/gx79r1u49o76w9ryb7dmoo9j5upmsao470.png)
![\Rightarrow\ \alpha=0.05](https://img.qammunity.org/2020/formulas/mathematics/college/sml8l4m8evuy9se82gksbjn4yqm41e5t2u.png)
Then, critical z-value =
![z_(\alpha/2)=1.96](https://img.qammunity.org/2020/formulas/mathematics/high-school/fn1e1isyr7r4ubq2yxfnpgs4mo3eo8m7ik.png)
The confidence interval to estimate the population mean is given by :_
![\overline{x}\ \pm\ z_(\alpha/2)(\sigma)/(√(n))](https://img.qammunity.org/2020/formulas/mathematics/high-school/dhcv28zw9jrfif2fy7sq1kyi2b71a9ipfg.png)
![=200\ \pm\ (1.96)(48)/(√(64))\\\\=200\pm11.76=(200-11.76, 200+11.76)=(188.24,\ 211.76 )](https://img.qammunity.org/2020/formulas/mathematics/college/2p0x4s52sqnr30xc49cjmkotoven2eq37u.png)
Hence, the 95% confidence interval to estimate the population mean can be computed as
.