Determine how many branches have a sales of:

Question

The average net sales / year of the products in 60 branches (N) of Ray. Cañero co. inc. is 85M (mean), With a standard deviation of 15M. 60 to 85M? 70 to 105M? 90 to 110M? Above 110M? Below 60M?

Answer 1

Hi Freiya,

μ = 85 , σ =15

(a) Z=(X-μ)/σ; Z=(60-85)/15 = -1.6667, Z =(85-85)/15=0

P(60 <  X < 85) = P(-1.6667< Z < 0)

                         = P(Z < 0) - P(Z ≤ -1.66)

                         = 0.5000 - 0.04846

                         = 0.4515

Number of branches = 0.4515*60 = 27

(b)  Z=(X-μ)/σ ; Z=(70-85)/15 = -1.000, Z =(105-85)/15=1.3333

P(70<X < 105) = P( -1.000 < Z < 1.333)

                        = P(Z < 1.3333) - P(Z < -1.000)

                        = 0.90824 -0. 15866

                        = 0.7495

Number of branches = 0.7495*60 = 45

(c)  Z=(X-μ)/σ ; Z=(90-85)/15 = 0.333, Z =(110-85)/15= 1.666

P(90 < X  <110)= P( 0.333< Z < 1.666)

                         = P(Z <1.6667) - P(Z < 0.3333)

                         = 0.95154 -  0.62930

                         =0.3222

Number of branches = 0.322*60 = 19

(d)Z=(X-μ)/σ ; Z=(110-85)/15= 1.6667 

 P(X  > 110) = P(Z > 1.6667)

                   = 1 - P(Z < 1.6667)

                   = 1 - 0.95154

                   = 0.04846

Number of branches = 0.04846*60 = 3

(e)Z=(X-μ)/σ ; Z=(60-85)/15 = -1.6667 

P(X  < 60) = P(Z < -1.6667)

                = 0.04846

Number of branches = 0.04846*60 = 3