stdev 2754668 average of sample data 6078947 Using the fol

stdev = 2.754668

average of sample data = 6.078947

Using the following guesses for = the mean number of hours spent shopping on Black Friday for all Black Friday shoppers to complete the table. Enter the z-scores calculated using the sample mean and the corresponding guess for .

mew = 5

mew = 6

mew = 7

mew = 8

is it right to do

(6.078947-5)/2.745668 =

(6.078947-6)/2.745668 =

(6.078947-7)/2.745668 =

(6.078947-8)/2.745668 =

Solution

Standard Deviation ( sd )= 2.754668
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
Mean ( u ) = 5

P(X = 6.078947) = (6.078947-5)/2.754668
= 1.0789/2.754668= 0.3917

b)
Mean ( u ) = 6

P(X = 6.078947) = (6.078947-6)/2.754668
= 0.0789/2.754668= 0.0287

c)
Mean ( u ) = 7

P(X < 6.078947) = (6.078947-7)/2.754668
= -0.9211/2.754668= -0.3344

d)
P(X < 6.078947) = (6.078947-8)/2.754668
= -1.9211/2.754668= -0.6974