The average amount customers at a certain grocery store spend yearly is $636.55. Assume the variable is normally distributed. If the standard deviation is $89.46, find the probability that a randomly selected customer spends between $550.67 and $836.94

Question

amistre64 · Answer

ah hah  stats

amistre64 · Answer

mean = 636.55
sd = 89.46

interval = 550.67 to 836.94

amistre64 · Answer

we need to squeeze this into the normal distribution curve by:

lower limit - mean;  upper limit - mean
---------------   ----------------
          sd                           sd

amistre64 · Answer

or; we can use our normalcdf(l,u,m,sd) on our ti83 :)

amistre64 · Answer

the ti83 gives me:  .81892...

anonymous · Answer

the answers are 0.144, 0.820, 0.156, or 0.943????

anonymous · Answer

so round it up?

amistre64 · Answer

id go .82 then ;)