Answer: a) 0.1587 b) 0.0618
Step-by-step explanation:
Let x be the random variable that represents the monthly demand for a product.
Given : The monthly demand for a product is normally distributed with mean = 700 and standard deviation = 200.
i.e. [tex]\mu=700[/tex] and [tex]\sigma=200[/tex]
a) Using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to x= 900 will be :
[tex]z=\dfrac{900-700}{200}=1[/tex]
Now, by using the standard normal z-table , the probability demand will exceed 900 units in a month :-
[tex]P(z>1)=1-P(z\leq1)=1-0.8413=0.1587[/tex]
Hence, the probability demand will exceed 900 units in a month=0.1587
a) Using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to x= 392 will be :
[tex]z=\dfrac{ 392-700}{200}=-1.54[/tex]
Now, by using the standard normal z-table , the probability demand will be less than 392 units in a month :-
[tex]P(z<-1.54)=1-P(z<1.54)=1-0.9382=0.0618[/tex]
Hence, the probability demand will be less than 392 units in a month = 0.0618
