Answer:
0.1056
Explanation:
A large company claims that the average age of their employees is 32 years i.e.
![\mu = 32](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4ghv4tinyy6e1urlno1cv65ivsd6o6ntbu.png)
Standard deviation =
![\sigma = 4](https://img.qammunity.org/2019/formulas/mathematics/middle-school/8k553j591cpt32zw3ogj5i334zk8bvu13l.png)
The average age of employees in the sales department at the company is 27 years i.e.
![\bar{x}=27](https://img.qammunity.org/2019/formulas/mathematics/middle-school/q7qbiwilin5biozk8modp2torae3ahhika.png)
The z-score corresponding to x = 27 is calculated as:
,
![z =(27-32)/(4) = -1.25](https://img.qammunity.org/2019/formulas/mathematics/middle-school/3rk7ztsvex7dc4fwudlljyzejg23wcqcmf.png)
Now, From a z-table,
The probability that x < 27 corresponds to the probability that z < -1.25:
P(z < -1.25) = 0.1056
Hence the probability that an employee, chosen at random, will be younger than 27 years is 0.1056