Answer:
"0.0125" is the right solution.
Explanation:
The given values are:
Random sample,
n = 90
Claims,
p = 20%
or,
= 0.20
By using normal approximation, we get
⇒
![\mu = np](https://img.qammunity.org/2022/formulas/mathematics/college/zaapkupaonewp2mvvsa5xz9t5zqcsg1zkr.png)
On substituting the values, we get
⇒
![=90* 0.20](https://img.qammunity.org/2022/formulas/mathematics/college/y2cki6xhj08o30g3dgyrvcu7d7drrijszk.png)
⇒
![=18](https://img.qammunity.org/2022/formulas/mathematics/high-school/7dkz597hi1pmzqq9urz25p2radig8kwnir.png)
Now,
The standard deviation will be:
⇒
![\sigma=√(np(1-p))](https://img.qammunity.org/2022/formulas/mathematics/college/izdz4npmrmosvuuvj4edingd7qg92h5vpt.png)
On putting the above given values, we get
⇒
![=√(90* 0.20* (1-0.20))](https://img.qammunity.org/2022/formulas/mathematics/college/tyqqtbx9bu3c1qa9tmx64h7u7c8o7yqtup.png)
⇒
![=√(18* 0.8)](https://img.qammunity.org/2022/formulas/mathematics/college/94g9klmvz2h9vewkneibaynl5ls468llq4.png)
⇒
![=√(14.4)](https://img.qammunity.org/2022/formulas/mathematics/college/2mgre6qb4y73237p6vhj2an0n24ewq7bve.png)
⇒
![=3.7947](https://img.qammunity.org/2022/formulas/mathematics/college/khop1noxylajvvdp4wdjn6iab3oau0tb28.png)
hence,
By using the continuity correction or the z-table, we get
⇒
![P(x < 10) = P(x < 9.5)](https://img.qammunity.org/2022/formulas/mathematics/college/z8g1ckjjtehna0j591x8cn05rjggpx51fv.png)
⇒
![P(x < 10) = P((x-\mu)/(\sigma) -(9.5-18)/(3.7947) )](https://img.qammunity.org/2022/formulas/mathematics/college/jqfkvmqa8qrqdxoz9re9xn3u70yp8tlipq.png)
⇒
![P(x < 10) = P(Z < -2.24)](https://img.qammunity.org/2022/formulas/mathematics/college/l1lu7skrf9crca5o7j2zhmh4xst5ji9ar0.png)
From table,
⇒
![P(x < 10) = 0.0125](https://img.qammunity.org/2022/formulas/mathematics/college/xbcmjdpmaqad38ibxmzd90lbm7xt0632vf.png)