Question

The weight of bags of fertilizer is normally distributed with a mean of 50 pounds and standard deviation of 6 pounds. What is the probability that a bag of fertilizer will weigh:

a. Between 45 and 55 pounds?
b. At least 56 pounds?
c. At most 49 pound?

Answer 1

Following are the solution to the given points:

Explanation:

Normal Distribution:

`\mu=50\\\\\sigma= 6\\\\Z=(X-\mu)/(\sigma) \sim N(O,l)`

For point a:

`P(X< 56)=((56-50))/(6)= (6)/(6)=1\\\\`

`=P(Z<1)\ From\ \sigma \ Table=0.8413\\\\P(X>= 56)=(1-P(X< 56))=1-0.8413=0.1587\\\\`

For point b:

`P(X< 49)=((49-50))/(6)=-(1)/(6) =-0.1667\\\\=P(Z<-0.1667)\ From\ \sigma \ Table\\\\=0.4338`

For point c:

To Find
`P(a\leq Z\leq b)= F(b) - F(a)\\\\`

`P(X< 45)=((45-50))/(6)=(-5)/(6) =-0.8333\\\\P (Z<-0.8333) \ From \ \sigma \ Table\\\\=0.20233\\\\P(X< 55)=((55-50))/(6) =(5)/(6)=0.8333\\\\P ( Z< 0.8333) \ From \ \sigma\ Table\\\\=0.79767\\\\P(45 < X < 55) =0.79767-0.20233 =0.5953`