Chapter 5, Problem 5.82SE

(a)

To determine

To calculate the probability that more than six consumers say they have become conservative spenders.

Answer to Problem 5.82SE

The probabilities that more than six consumers say they have become conservative spenders is $0.9050$ .

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=60\%=0.6\end{array}$

Now, the probability that more than six consumers say they have become conservative spenders implies,

  $P(X>6)=1-P(X\le 6)$

Where,

  $\begin{array}{l}P(X\le 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ +P(X=4)+P(X=5)+P(X=6)\end{array}$

Which implies,

  $P(X=0)=C_0^{15}.{(0.6)}^0.{(1-0.6)}^{15-0}$

  $=\frac{15!}{0!(15-0)!}\times {(0.6)}^0\times {(1-0.6)}^{15-0}$

  $=0.0000$

  $P(X=1)=C_1^{15}.{(0.6)}^1.{(1-0.6)}^{15-1}$

  $=\frac{15!}{1!(15-1)!}\times {(0.6)}^1\times {(1-0.6)}^{14}$

  $=0.0000$

  $P(X=2)=C_2^{15}.{(0.6)}^2.{(1-0.6)}^{15-2}$

  $=\frac{15!}{2!(15-2)!}\times {(0.6)}^2\times {(1-0.6)}^{13}$

  $=0.0003$

  $P(X=3)=C_3^{15}.{(0.6)}^3.{(1-0.6)}^{15-3}$

  $=\frac{15!}{3!(15-3)!}\times {(0.6)}^3\times {(1-0.6)}^{12}$

  $=0.0016$

  $P(X=4)=C_4^{15}.{(0.6)}^4.{(1-0.6)}^{15-4}$

  $=\frac{15!}{4!(15-4)!}\times {(0.6)}^4\times {(1-0.6)}^{11}$

  $=0.0074$

  $P(X=5)=C_5^{15}.{(0.6)}^5.{(1-0.6)}^{15-5}$

  $=\frac{15!}{5!(15-5)!}\times {(0.6)}^5\times {(1-0.6)}^{10}$

  $=0.0245$

  $P(X=6)=C_6^{15}.{(0.6)}^6.{(1-0.6)}^{15-6}$

  $=\frac{15!}{6!(15-6)!}\times {(0.6)}^6\times {(1-0.6)}^9$

  $=0.0612$

Thus, we have,

  $\begin{array}{l}P(X\le 6)=0.0000+0.0000+0.0003+0.0016\\ +0.0074+0.0245+0.0612\\ \Rightarrow P(X\le 6)=0.0950\end{array}$

And now,

  $\begin{array}{l}P(X>6)=1-P(X\le 6)\\ =1-0.0950\\ =0.9050\end{array}$

(b)

To determine

To calculate the probability that fewer than five of those sampled have become conservative spenders.

Answer to Problem 5.82SE

The probabilities that fewer than five of those sampled have become conservative spendersis $0.0093$ .

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=60\%=0.6\end{array}$

Now, the probability that fewer than five of those sampled have become conservative spenders implies,

  $\begin{array}{l}P(X<5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ +P(X=4)\end{array}$

Which implies,

  $\begin{array}{}$ P(X=0)=C_0^{15}.{(0.6)}^0.{(1-0.6)}^{15-0}$$ =\frac{15!}{0!(15-0)!}\times {(0.6)}^0\times {(1-0.6)}^{15-0}$$ =0.0000$$ P(X=1)=C_1^{15}.{(0.6)}^1.{(1-0.6)}^{15-1}$$ =\frac{15!}{1!(15-1)!}\times {(0.6)}^1\times {(1-0.6)}^{14}$$ =0.0000$$ P(X=2)=C_2^{15}.{(0.6)}^2.{(1-0.6)}^{15-2}$$ =\frac{15!}{2!(15-2)!}\times {(0.6)}^2\times {(1-0.6)}^{13}$$ =0.0003$$ P(X=3)=C_3^{15}.{(0.6)}^3.{(1-0.6)}^{15-3}$$ =\frac{15!}{3!(15-3)!}\times {(0.6)}^3\times {(1-0.6)}^{12}$$ =0.0016$$ P(X=4)=C_4^{15}.{(0.6)}^4.{(1-0.6)}^{15-4}$$ =\frac{15!}{4!(15-4)!}\times {(0.6)}^4\times {(1-0.6)}^{11}$$ =0.0074$\end{array}$

Thus, we have,

  $\begin{array}{l}P(X<5)=0.0000+0.0000+0.0003+0.0016+0.0074\\ \Rightarrow P(X<5)=0.0093\end{array}$

(c)

To determine

To calculate the probability that exactly nine of those sampled have become conservative spenders.

Answer to Problem 5.82SE

The probabilities that exactly nine of thosesampled have become conservative spenders is $0.2066$ .

Explanation of Solution

Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use the Binomial distribution.

The binomial probability is defined as,

  $P(X=k)=C_k^n.p^k.q^{n-k}=\frac{n!}{k!(n-k)!}\times p^k\times {(1-p)}^{n-k}$

Where $\begin{array}{l}n=\text{ number of trials }=15\\ p=\text{ Probability of success }=60\%=0.6\end{array}$

Now, the probability that exactly nine of thosesampled have become conservative spenders implies,

  $\begin{array}{l}$ P(X=9)=C_9^{15}.{(0.6)}^9.{(1-0.6)}^{15-9}$$ =\frac{15!}{9!(15-9)!}\times {(0.6)}^9\times {(1-0.6)}^6$=0.2066\end{array}$