a.
If X is the daily sales
then X follows normal distribution with a mean = 150000 and Std deviation= 15000
i. No of branches with sales between 120000 & 145000
When X= 120000, Z= (X-mean)/Standard deviation
X =120000 : Z = -2
X= 145000 :Z = -0.33
Now, we calculate P(-2<=Z<=-0.33)= P(0.33<=Z<=2)=P(Z<=2)-P(Z<=0.33)=0.97725-0.62930=0.34795 (From Normal Tables)
Multiplying the probability with 500 to get the number of offices = 500*0.34795= 173.975 or 174 offices approximately.
ii.140000 & 165000
Applying a similar procedure:
X= 140000: Z= -0.66
X= 165000: Z= 1
Now, we calculate P(-0.66<=Z<=1)= P(-0.66<=Z<=0)+P(0<=Z<=1)=0.24537+0.34134=0.58671 (From Normal Tables)
Multiplying the probability with 500 to get the number of offices = 500*0.58671= 293.355 or 293 offices approximately.
a) The average daily sales of 500 branch offices was Ksh 150000 with a standard deviation...