The daily sales at a convenience store produce a distribution that is approximat

Question

The daily sales at a convenience store produce a distribution that is approximately normal with a mean of 1160 and a standard deviation of 149. In your intermediate calculations, round z-values to two decimal places. The probability that the sales on a given day at this store are more than $ 1405, rounded to four decimal places, is: . The probability that the sales on a given day at this store are less than $ 1305, rounded to four decimal places, is: . The probability that the sales on a given day at this store are between $ 1200 and $ 1300, rounded to four decimal places, is: .

Explanation / Answer

Mean ( u ) =1160
Standard Deviation ( sd )=149
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X > 1405) = (1405-1160)/149
= 245/149 = 1.6443
= P ( Z >1.644) From Standard Normal Table
= 0.0501                  
b.
P(X < 1305) = (1305-1160)/149
= 145/149= 0.9732
= P ( Z <0.9732) From Standard Normal Table
= 0.8348                  
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1200) = (1200-1160)/149
= 40/149 = 0.2685
= P ( Z <0.2685) From Standard Normal Table
= 0.60583
P(X < 1300) = (1300-1160)/149
= 140/149 = 0.9396
= P ( Z <0.9396) From Standard Normal Table
= 0.82629
P(1200 < X < 1300) = 0.82629-0.60583 = 0.2205                  

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