Answer:
The confidence interval is
![13.4< \mu < 13.8](https://img.qammunity.org/2021/formulas/mathematics/college/82z2bbotvx9s1dqo33plmrj7axrm2rqrk3.png)
Explanation:
From the question we are told that
The sample mean is
![\= x = 13.6](https://img.qammunity.org/2021/formulas/mathematics/college/iysvqqeobaayx3u5jpzlm74mrc7jvt5x3m.png)
The standard deviation is
![\sigma = 1.9](https://img.qammunity.org/2021/formulas/mathematics/college/1dt03j1jnbdsx7qhxqpmo9t9wrzeob4o7d.png)
The sample size is
![n = 189](https://img.qammunity.org/2021/formulas/mathematics/college/b4iiqsfm8c5mwytqq2nbop7uvtsfogcml2.png)
given that the confidence level is 85% then the level of significance is mathematically represented as
![\alpha = 0.15](https://img.qammunity.org/2021/formulas/mathematics/college/k9gwejgcuq5h3shafqjn56bv8ywc5tj3o0.png)
Next we obtain the critical value of
from the normal distribution table the value is
![Z_{(\alpha )/(2) } = 1.44](https://img.qammunity.org/2021/formulas/mathematics/college/m98ez41vg9jzgvstdmzlfi5u3825kquedm.png)
Generally the margin of error is mathematically represented as
![E = Z_{(\alpha )/(2) } * (\sigma)/(√(n) )](https://img.qammunity.org/2021/formulas/mathematics/college/tv87b94ol5pfn8u7m0cjz5lbcv6yrhnoau.png)
=>
![E = 1.44* (1.9)/(√(189) )](https://img.qammunity.org/2021/formulas/mathematics/college/prbsff1nl36fksgy6nvnu865urp9hex55d.png)
=>
![E = 0.1990](https://img.qammunity.org/2021/formulas/mathematics/college/ck1p5rys9ftd7qb9jmgifgdlosuke9l479.png)
The 85% confidence interval is mathematically represented as
![\= x - E < \mu <\= x + E](https://img.qammunity.org/2021/formulas/mathematics/college/5azdicutkilelv2z4716ggppi3xafu5o58.png)
=>
![13.6- 0.1990 < \mu < 13.6+ 0.1990](https://img.qammunity.org/2021/formulas/mathematics/college/iki3laanq9f0u5kvao6bby0trqrn8avd6m.png)
=>
![13.4< \mu < 13.8](https://img.qammunity.org/2021/formulas/mathematics/college/82z2bbotvx9s1dqo33plmrj7axrm2rqrk3.png)