Question

A company that supplies ready-mix concrete receives, on average, six orders per day.

(a) What is the probability that, on a given day:

(i) only one order will be received?

(ii) no more than three orders will be received?

(iii) at least three orders will be received?

(b) What is the probability that, on a given half-day, only one order will be received?

(c) What is the mean and standard deviation of orders received per day?

Answer 1

Solution:

Let count all possible outcomes and ways:.

Number of possible outcomes: $N=2^{n};$ here $n$ - number of orders.

Number of ways: $^{n}C_{m}=\frac{n!}{(n-m)!m!};$ here $m$ - number of received orders. So,

$^{6}C_{0}=\frac{6!}{(6-0)!0!}=1;$

$^{6}C_{1}=\frac{6!}{(6-1)!1!}=6;$

$^{6}C_{2}=\frac{6!}{(6-2)!2!}=15;$

$^{6}C_{3}=\frac{6!}{(6-3)!3!}=20;$

$^{6}C_{4}=\frac{6!}{(6-4)!4!}=15;$

$^{6}C_{5}=\frac{6!}{(6-5)!5!}=6;$

$^{6}C_{6}=\frac{6!}{(6-6)!6!}=1;$

$N=2^{6}=64;$

a)

1) $P(1)=\frac{6}{64}=0.09375;$

2) No more than three, so, we need to add terms with m from 0 to 3:

$P(m\le3)=P(0)+P(1)+P(2)+P(3)=\frac{1+6+15+20}{64}=0.65625;$

3) At least three order received, so, we need to add terms with m from 3 to 6;

$P(m\ge3)=P(3)+P(4)+P(5)+P(6)=\frac{20+15+6+1}{64}=0.65625;$

b) If a half day, it means $n=\frac{6}{2}=3;$

$N=2^{3}=8-$ numbers of possible outcomes;

Number of ways:

$^{3}C_{0}=\frac{3!}{(3-0)!0!}=1;$

$^{3}C_{1}=\frac{3!}{(3-1)!1!}=3;$

$^{3}C_{2}=\frac{3!}{(3-2)!2!}=3;$

$^{3}C_{3}=\frac{3!}{(3-3)!3!}=1;$

$P(1)=\frac{3}{8}=0.375;$

c) In the binomial distribution:

$\mu=nP(1)=64\times0.09375=6$ is the mean of the distribution;

$\sigma=\sqrt{nP(1)(1-P(1))}=2.33$ is the standard deviation of the distribution.

Answer:

a)

1) $P(1)=0.09375;$

2) $P(m\le3)=0.65625;$

3) $P(m\ge3)=0.65625;$

b) $P(1)=0.375;$

c) $\mu=6;$ $\sigma=2.33.$