A bank has 1500 individual savings accounts with an average

A bank has 1,500 individual savings accounts with an average balance of $3,000 and a standard deviation of $1,200. If the bank takes a random sample of 100 accounts,

(a) what\'s the probability that the average savings for these 100 accounts will be below $2,800?

(b) how many accounts are expected to be below $2,800?

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2800      
u = mean =    3000      
n = sample size =    100      
s = standard deviation =    1200      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.666666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.666666667   ) =    0.047790352 [answer]

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b)

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2800      
u = mean =    3000      
n = sample size =    100      
s = standard deviation =    1200      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.666666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.666666667   ) =    0.047790352 [answer]

********************

b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2800      
u = mean =    3000      
          
s = standard deviation =    1200      
          
Thus,          
          
z = (x - u) / s =    -0.166666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.166666667   ) =    0.433816167

Thus, we expect 100*0.433816167 = 43.3816167 to be below 2800. [ANSWER, 43.3816167]