Question

Suppose that it is known that the number of items produced in a factory during a week is a random variable with mean 50. If the variance of a weekâs production is known to equal 24, then what can be said about the probability that this weekâs production will be between 40 and 60?

Answer 1

Answer:

The probability would be 0.9586

Step-by-step explanation:

Given,

Mean, $\mu = 50$,

Variance, $\sigma^2 = 24$

Standard deviation,

$\sigma = \sqrt{24}=2\sqrt{6}$,

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

$=P(40 < X < 60)$

$=P(\frac{40-\mu}{\sigma} < \frac{X-\mu}{\sigma} < \frac{60-\mu}{\sigma})$

$=P(\frac{40-50}{2\sqrt{6}} < z < \frac{60-50}{2\sqrt{6}})$

$=P(-\frac{10}{2\sqrt{6}} < z < \frac{10}{2\sqrt{6}})$

$=P(-2.041 < z < 2.041)$

$=P(z< 2.041) - P(z< -2.041)$

Using z score table

= 0.9793 - 0.0207

= 0.9586