Chapter 2.2, Problem 72E

To determine

Mean and standard deviation of the distribution.

Answer to Problem 72E

Mean,

  $\mu =1.1087$

Standard deviation,

  $\sigma =0.4348$

Explanation of Solution

Given information:

Ricardo spends less than 1 minute brushing his teeth about 40% of the time.

Ricardo spends more than 2 minutes brushing his teeth 2% of the time.

Calculations:

Ricardo spends less than 1 minute about 40% of the time.

Now,

Find the z − score corresponding the probability of 40% (or 0.40) in the normal probability table of the appendix.

Note that

The closest probability would be 0.40129 that lies in the row -0.2 and in the column .05 of the normal probability table.

Then

The corresponding z − score,

  $z=-0.25$

And

Ricardo spends more than 2 minutes about 2% of the time.

That means

Ricardo spends less than 2 minutes about 98% of the time.

Now,

Find the z − score corresponding the probability of 98% (or 0.98) in the normal probability table of the appendix.

Note that

The closest probability would be 0.9798 that lies in the row 2.0 and in the column .05 of the normal probability table.

Then

The corresponding z − score,

  $z=2.05$

Now,

The value will be the mean increased by the product of the z − value and the $\sigma$ (standard deviation).

  $1=x=\mu +z\sigma =\mu -0.25\sigma$

And

  $2=x=\mu +z\sigma =\mu +2.05\sigma$

Subtract the above two equations:

  $1=2.30\sigma$

Divide both sides by 2.30:

We have

Standard deviation,

  $\sigma =\frac{1}{2.30}=0.4348$

Then

Find the mean:

  $\begin{array}{l}\mu =1+0.25\sigma \\ =1+0.25\left(0.4348\right)\\ =1.1087\end{array}$