Chapter 5.2, Problem 22P

a.

To determine

Obtain a trial for the given problem.

Describe a success for the given problem.

Describe a failure for the given problem.

Mention the values for n, p, and q.

Compute the probability of grossing the store over $850 on at least 3 out of 5 business days.

Answer to Problem 22P

A trial is gross receipt of store for one business day.

The success is described as the store grossed over $850.

The failure is described as the store grossed $850 or below.

Total number of business days (or trials) is $n=5$.

The probability that the store grossed over $850 is $p=0.60$.

The probability that the store grossed $850 or below is $q=0.40$.

The probability of grossing the store over $850 on at least 3 out of 5 business days is 0.683.

Explanation of Solution

Calculation:

Trial:

A trial is gross receipt of store for one business day with two possible outcomes success or failure.

Success:

Consider success as the store grossed over $850.

Failure:

Consider failure as the store grossed $850 or below.

Values:

Total number of business days (or trials) is $n=5$.

The probability that the store grossed over $850 (or success) is $p=0.60$.

The probability that the store grossed $850 or below (or failure) is calculated as given below:

$\begin{array}{c}q=1-p\\ =1-0.60\\ =0.40\end{array}$

Random Variable:

Let r be a binomial random variable, which represents the number of business days that the stores gross over $850.

Binomial probability:

The probability of r successes out of n trials is given below:

$P(r)=C_{n,r}{p}^r{q}^{n-r}$

Here, n is the number of trials, r is the number of successes, p is the probability of success, and q is the probability of failure.

The probability that the store gross over $850 on at least 3 out of 5 business days is calculated as given below:

$\begin{array}{l}P(r\ge 3)=P\left(3\right)+P\left(4\right)+P\left(5\right)\\ =C_{5,3}{\left(0.60\right)}^3{\left(0.40\right)}^2+C_{5,4}{\left(0.60\right)}^4{\left(0.40\right)}^1+C_{5,5}{\left(0.60\right)}^5{\left(0.40\right)}^0\\ =\left(\frac{5!}{3!2!}\times {\left(0.60\right)}^3\times {\left(0.40\right)}^2\right)+\left(\frac{5!}{4!1!}\times {\left(0.60\right)}^4\times {\left(0.40\right)}^1\right)+\left(\frac{5!}{5!0!}\times {\left(0.60\right)}^5\times {\left(0.40\right)}^0\right)\\ =0.346+0.259+0.078\\ =0.683\end{array}$

Therefore, the probability that the store gross over $850 on at least 3 out of 5 business days is 0.683.

b.

To determine

Compute the probability of grossing the store over $850 on at least 6 out of 10 business days.

Answer to Problem 22P

The probability of grossing the store over $850 on at least 6 out of 10 business days is 0.633.

Explanation of Solution

Calculation:

Total number of business days (or trials) is $n=10$.

The probability that the store gross over $850 on at least 6 out of 10 business days is calculated as given below:

$\begin{array}{l}P(r\ge 6)=P\left(6\right)+P\left(7\right)+P\left(8\right)+P\left(9\right)+P\left(10\right)\\ =C_{10,6}{\left(0.60\right)}^6{\left(0.40\right)}^4+C_{10,7}{\left(0.60\right)}^7{\left(0.40\right)}^3+C_{10,8}{\left(0.60\right)}^8{\left(0.40\right)}^2+C_{10,9}{\left(0.60\right)}^9{\left(0.40\right)}^1\\ +C_{10,10}{\left(0.60\right)}^{10}{\left(0.40\right)}^0\\ =0.251+0.215+0.121+0.040+0.006\\ =0.633\end{array}$

Therefore, the probability that the store gross over $850 on at least 6 out of 10 business days is 0.633.

c.

To determine

Compute the probability of grossing the store over $850 on less than 5 out of 10 business days.

Answer to Problem 22P

The probability of grossing the store over $850 on less than 5 out of 10 business days is 0.166.

Explanation of Solution

Calculation:

Total number of business days (or trials) is $n=10$.

The probability that the store gross over $850 on less than 5 out of 10 business days is calculated as given below:

$\begin{array}{l}P(r<5)=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)\\ =C_{10,0}{\left(0.60\right)}^0{\left(0.40\right)}^{10}+C_{10,1}{\left(0.60\right)}^1{\left(0.40\right)}^9+C_{10,2}{\left(0.60\right)}^2{\left(0.40\right)}^8+C_{10,3}{\left(0.60\right)}^3{\left(0.40\right)}^7\\ +C_{10,4}{\left(0.60\right)}^4{\left(0.40\right)}^6\\ =0.000+0.002+0.011+0.042+0.111\\ =0.166\end{array}$

Therefore, the probability that the store gross over $850 on less than 5 out of 10 business days is 0.166.

d.

To determine

Compute the probability of grossing the store over $850 on less than 6 out of 20 business days.

Answer to Problem 22P

The probability of grossing the store over $850 on less than 6 out of 20 business days is 0.001.

Explanation of Solution

Calculation:

Total number of business days (or trials) is $n=20$.

The probability that the store gross over $850 on less than 6 out of 20 business days is calculated as given below:

$\begin{array}{l}P(r<6)=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P(5)\\ =C_{20,0}{\left(0.60\right)}^0{\left(0.40\right)}^{20}+C_{20,1}{\left(0.60\right)}^1{\left(0.40\right)}^{19}+C_{20,2}{\left(0.60\right)}^2{\left(0.40\right)}^{18}+C_{20,3}{\left(0.60\right)}^3{\left(0.40\right)}^{17}\\ +C_{20,4}{\left(0.60\right)}^4{\left(0.40\right)}^{16}+C_{20,5}{\left(0.60\right)}^5{\left(0.40\right)}^{15}\\ =0.000+0.000+0.000+0.000+0.000+0.001\\ =0.001\end{array}$

Therefore, the probability that the store gross over $850 on less than 6 out of 20 business days is 0.001.

Interpretation:

The event of 20 business days with gross income over $850 on less than 6 days is rare if the probability of success is 0.60. If it happened again, then high probability of success $p=0.60$ is suspected.

e.

To determine

Compute the probability of grossing the store over $850 on more than 17 out of 20 business days.

Answer to Problem 22P

The probability of grossing the store over $850 on more than 17 out of 20 business days is 0.003.

Explanation of Solution

Calculation:

Total number of business days (or trials) is $n=20$.

The probability that the store gross over $850 on more than 17 out of 20 business days is calculated as given below:

$\begin{array}{l}P(r>17)=P\left(18\right)+P\left(19\right)+P(20)\\ =C_{20,18}{\left(0.60\right)}^{18}{\left(0.40\right)}^2+C_{20,19}{\left(0.60\right)}^{19}{\left(0.40\right)}^1+C_{20,20}{\left(0.60\right)}^{20}{\left(0.40\right)}^0\\ =0.003+0.000+0.000\\ =0.003\end{array}$

Therefore, the probability that the store gross over $850 on more than 17 out of 20 business days is 0.003.

Interpretation:

The event of 20 business days with gross income over $850 on more than 17 days is rare if the probability of success is 0.60. If it happens again, then low probability of success $p=0.60$ is suspected.