Question

A specific drive through receives an average of 32 customers per 12 hours with the standard deviation of 5. What is the probability that they receive less than 20 customers in the next 12 hours?

Answer 1

Solution :

Given, X follows Normal distribution with,

  $\mu$ = 32

  $\sigma$ = 5

Find P(X < 20)

= P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596147763&ymreqid=69d777f6-ff7f-1f97-1cad-0c004c01aa00&sig=kExP9Gc_ea90PpaRBoDpWA--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596147763&ymreqid=69d777f6-ff7f-1f97-1cad-0c004c01aa00&sig=lLV94Hhfh8BZ6mIVI6ighw--~D$ <  (20 - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596147763&ymreqid=69d777f6-ff7f-1f97-1cad-0c004c01aa00&sig=kExP9Gc_ea90PpaRBoDpWA--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596147763&ymreqid=69d777f6-ff7f-1f97-1cad-0c004c01aa00&sig=lLV94Hhfh8BZ6mIVI6ighw--~D$]

= P[Z <  (20 - 32)/5]

= P[Z < -2.4]

= 0.0082 ... ( use z table)

P(X < 20) = 0.0082

Answer : The probability that they receive less than 20 is 0.0082