91Ó°ÊÓ

It is given that 70% of the adults always wash their hands after using a public restroom.

a.

Let X denote the number of adults whoalways wash their hands after using a public restroom.

Let success be defined as selecting an adult whoalways washeshis/her hands after using a public restroom.

The number of trials (n) is given to be equal to 8.

The probability of success is equal to

\(\begin{aligned}{c}p = 70\% \\ = \frac{{70}}{{100}}\\ = 0.70\end{aligned}\)

The probability of failure is equal to

\(\begin{aligned}{c}q = 1 - p\\ = 1 - 0.70\\ = 0.30\end{aligned}\)

The number of successes required in 8 trials should be equal to x=5.

The binomial probability formula is as follows:

\(P\left( {X = x} \right) = {\;^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}\)

Using the binomial probability formula, the probability of getting exactly 5 adults whoalways wash their hands after using a public restroomis computed below:

\(\begin{aligned}{c}P\left( {X = 5} \right) = {\;^8}{C_5}{\left( {0.70} \right)^5}{\left( {0.30} \right)^{8 - 5}}\\ = \frac{{8!}}{{5!\left( {8 - 5} \right)!}}{\left( {0.70} \right)^5}{\left( {0.30} \right)^3}\\ = 0.254\end{aligned}\)

Therefore, the probability of getting exactly 5 adults whoalways wash their hands after using a public restroomisequal to 0.254.

b.

Using the binomial probability formula, the probability that at least 7 out of 8 adultsalways wash their hands after using a public restroomis computed below:

\(\begin{aligned}{c}P\left( {X \ge 7} \right) = P\left( {X = 7} \right) + P\left( {X = 8} \right)\\ = {\;^8}{C_7}{\left( {0.70} \right)^7}{\left( {0.30} \right)^{8 - 7}} + {\;^8}{C_8}{\left( {0.70} \right)^8}{\left( {0.30} \right)^{8 - 8}}\\ = 0.1976 + 0.0576\\ = 0.2552\end{aligned}\)

\( \approx 0.255\)

Therefore, the probability that at least 7 adultsalways wash their hands after using a public restroom is equal to 0.255.

c.

The mean number of adults whoalways wash their hands after using a public restroomis equal to:

\(\begin{aligned}{c}\mu = np\\ = \left( 8 \right)\left( {0.70} \right)\\ = 5.6\end{aligned}\)

Therefore, the mean number of adults whoalways wash their hands after using a public restroomis equal to 5.6 adults.

The standard deviation is computed below:

\(\begin{aligned}{c}\sigma = \sqrt {npq} \\ = \sqrt {\left( 8 \right)\left( {0.70} \right)\left( {0.30} \right)} \\ = 1.3\end{aligned}\)

Therefore, the standard deviation of the number of adults who always wash their hands after using a public restroom is equal to 1.3.

d

Using the range rule of thumb, the significantly low number of adults whoalways wash their hands after using a public restroomis equal to:

\(\begin{aligned}{c}\mu - 2\sigma = 5.6 - \left( 2 \right)\left( {1.3} \right)\\ = 3.0\end{aligned}\)

Thus, a significantly low number of adults whoalways wash their hands after using a public restroomis less than or equal to 3.0.

Significantly high number of adults whoalways wash their hands after using a public restroomis equal to:

\(\begin{aligned}{c}\mu + 2\sigma = 5.6 + \left( 2 \right)\left( {1.3} \right)\\ = 8.2\end{aligned}\)

Thus, significantly high number of adults whoalways wash their hands after using a public restroomis greater than or equal to 8.2.

And the values that are not significant will lie between 3.0 and 8.2.

Here, the value of 1 is less than 3.0.

Therefore, the value of 1 adultwhoalways washes hands after using a public restroom is considered significantly low as it is less than 3.0.