1 The table below shows the probability distribution of the

(1) The table below shows the probability distribution of the demand for bread (in hundreds) of loaves per day at a bakery in a large metropolitan city:

X (lovaes in hundreds)

       P(X)

        1

      0.06

       2

     0.30

       3

      .40

       4

.      20

        5

      0.04

What is the expected or mean demand (in hundreds) and standard deviation for bread?

What is the probability that the demand for bread is less than 300 hundred loaves per day?

(2) Waiting time at a bank follows a normal distribution with mean ?=10 minutes and standard deviation

?=2 minutes.

Find P(X > 15).

(3) One would expect 2.5% of customers in the above problem to wait ____ minutes or more at the bank.

(4) Consider the standard normal distribution and find P(-1.45 < Z < 2.35).

Solution

The table below shows the probability distribution of the demand for bread (in hundreds) of loaves per day at a bakery in a large metropolitan city:

X (lovaes in hundreds)

       P(X)

        1

      0.06

       2

     0.30

       3

      .40

       4

.      20

        5

      0.04

What is the expected or mean demand (in hundreds) and standard deviation for bread?

expected or mean demand = 1*0.06+2*0.30+3*0.40+4*0.20+5*0.06

   = 2.96 (in hundreds)

What is the probability that the demand for bread is less than 300 hundred loaves per day?

probability that the demand for bread is less than 300 hundred loaves per day

=0.30+0.06 = 0.36

(2) Waiting time at a bank follows a normal distribution with mean ?=10 minutes and standard deviation ?=2 minutes.

Find P(X > 15).

Z value corresponding to 15 = (15-10)/ 2 = 2.5

P( X >15 ) = P(z > 2.5) = 0.0062

Z value corresponding 2.5% upper side =1.96

The required X value = 10+1.96*2 = 13.92

(4) Consider the standard normal distribution and find P(-1.45 < Z < 2.35).

P(-1.45 < Z < 2.35).

= P( Z < 2.35).- P(Z < -1.45 ). = 0.9906 - 0.0735 = 0.9171