Question

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of

12611261

chips and a standard deviation of

118118

chips. ​(a) Determine the

2525th

percentile for the number of chocolate chips in a bag. ​(b) Determine the number of chocolate chips in a bag that make up the middle

9595​%

of bags.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

​(a) The

2525th

percentile for the number of chocolate chips in a bag of chocolate chip cookies is

11821182

chocolate chips.

​(Round to the nearest whole number as​ needed.)

​(b) The number of chocolate chips in a bag that make up the middle

9595​%

of bags is

14831483

to

15621562

chocolate chips.

​(Round to the nearest whole number as needed. Use ascending​ order.)

Answer for all parts please.

Answer 1

a)

µ=   1261                  
σ =    118                  
proportion=   0.250                  
                      
Z value at    0.25   =   -0.6745   (excel formula =NORMSINV(   0.25   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.6745   *   118   +   1261  
X   =   1181.41              

25th percentile is 1182

b)

µ =    1261                          
σ =    118                          
proportion=   0.95                          
proportion left    0.05   is equally distributed both left and right side of normal curve                       
z value at   0.025   = ±   1.960   (excel formula =NORMSINV(   0.05   / 2 ) )      
                              
z = ( x - µ ) / σ                              
so, X = z σ + µ =                              
X1 =   -1.960   *   118   +   1261   =   1029.724  
X2 =   1.960   *   118   +   1261   =   1492.276  

the number of chocolate chips in a bag that make up the middle

95​%

of bags is (1030 , 1493)

c)

µ=   1261                  
σ =    118                  
proportion=   0.750                  
                      
Z value at    0.75   =   0.6745   (excel formula =NORMSINV(   0.75   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   0.6745   *   118   +   1261  
X   =   1340.59              

interquartile range = 75th percentile - 25th percentile = 1340.59-1181.41 = 159.18~159

please revert for doubts