Answer:
0.9168
Explanation:
From the data given:
Mean = 110
standard deviation = 5
Let consider a random sample n =49 which have a mean between 109 and 112.
The test statistics can be computed as:
![Z_1 = (x- \bar x)/((\sigma)/(√(n)))](https://img.qammunity.org/2021/formulas/mathematics/high-school/xn2ycd7yhfktgn3ti6xdnglyb418shstdo.png)
![Z_1 = (109- 110)/((5)/(√(49)))](https://img.qammunity.org/2021/formulas/mathematics/high-school/32v5ttgbm7fbw93jnsltyjc28d72we1x70.png)
![Z_1= (-1)/((5)/(7))](https://img.qammunity.org/2021/formulas/mathematics/high-school/2k67uti499f5cbomnjstr7e9uvbohxwql9.png)
= -1.4
![Z_2= (x- \bar x)/((\sigma)/(√(n)))](https://img.qammunity.org/2021/formulas/mathematics/high-school/tv61qer89s5yh299ux6u70pccsohrfhjk1.png)
![Z_2 = (112- 110)/((5)/(√(49)))](https://img.qammunity.org/2021/formulas/mathematics/high-school/hl90ipxm1bcxzr01pix3s1w74da7r03wsz.png)
![Z_2 = (2)/((5)/(7))](https://img.qammunity.org/2021/formulas/mathematics/high-school/4mq7h51tfbitk7bk8pdymdwsxjtpkaqut8.png)
![Z_2 =2.8](https://img.qammunity.org/2021/formulas/mathematics/high-school/n3vjhj4ecls5edlnr4zdw9x03fgdxooq7c.png)
Thus; P(109 <
< 112) = P( - 1.4 < Z < 2.8)
= P(Z < 2.8) - P( Z < -1.4)
= 0.9974 - 0.0806
= 0.9168