Answer:
The probability would be 0.9586
Step-by-step explanation:
Given,
Mean, [tex]\mu = 50[/tex],
Variance, [tex]\sigma^2 = 24[/tex]
Standard deviation,
[tex]\sigma = \sqrt{24}=2\sqrt{6}[/tex],
If X represents the number of items produced in a factory during a week.
Hence, the probability that this weeks production will be between 40 and 60
[tex]=P(40 < X < 60)[/tex]
[tex]=P(\frac{40-\mu}{\sigma} < \frac{X-\mu}{\sigma} < \frac{60-\mu}{\sigma})[/tex]
[tex]=P(\frac{40-50}{2\sqrt{6}} < z < \frac{60-50}{2\sqrt{6}})[/tex]
[tex]=P(-\frac{10}{2\sqrt{6}} < z < \frac{10}{2\sqrt{6}})[/tex]
[tex]=P(-2.041 < z < 2.041)[/tex]
[tex]=P(z< 2.041) - P(z< -2.041)[/tex]
Using z score table
= 0.9793 - 0.0207
= 0.9586
