It is believed that on average households spend 150 a week o

It is believed that on average, households spend $150 a week on groceries with a standard deviation of $50. What is the probability that in a sample of 64 households, the average amount spent on groceries will be no more than $140 per week

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    140      
u = mean =    150      
n = sample size =    64      
s = standard deviation =    50      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.6      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.6   ) =    0.054799292 [ANSWER]