Almonds are packaged so the mean almond weight is 05 ounce a

Almonds are packaged so the mean almond weight is .05 ounce and the standard deviation is .015 ounce. The almond weights are distributed normally. Suppose a sample of size 64 will be taken and the sample mean will be calculated. What is the probability that the mean of the sample of almonds is between .048 and .053 ounce?

Select one:

a. .1323

b. .1979

c. .8021

d. .8677

e. Cannot be determined

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    0.048      
x2 = upper bound =    0.053      
u = mean =    0.05      
n = sample size =    64      
s = standard deviation =    0.015      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.066666667      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.6      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.143061192      
P(z < z2) =    0.945200708      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.802139516 = 0.8021 [ANSWER, OPTION C]