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 an almond 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) / s, then as          
x1 = lower bound =    0.048      
x2 = upper bound =    0.053      
u = mean =    0.05      
          
s = standard deviation =    0.015      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.133333333      
z2 = upper z score = (x2 - u) / s =    0.2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.446964883      
P(z < z2) =    0.579259709      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.132294826 = 0.1323 [answer, OPTION A]