Chapter 2.2, Problem 62E

To determine

the mean and standard deviation of this distribution

Answer to Problem 62E

The required mean and standard deviation are 0.4348min and 1.1087min respectively.

Explanation of Solution

Given:

R less than one minute to brush the teeth 40% of the time

Spends more than two minutes brushing his teeth 2% of the time

Calculation:

It is given that the amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation.

Ricardo spends less than one minute brushing his teeth about 40% of the time. He spends more than two minutes brushing his teeth 2% of the time.

That is, P (brushing teeth for less than one minute) = 0.40 and the z value corresponding to 0.40 probability from standard normal table is $-0.2534$ .

Also, P (brushing teeth for more than two minutes) = 0.02. Then, brushing teeth for less than two minutes is 98% = 0.98. The z value corresponding to 0.98 probability from standard normal table is 2.054.

The following graph illustrates the given information to set up two equations in two unknowns.

  

Then, we can find the values of unknown mean and standard deviation using the following two equations as follows:

  $-0.25=\frac{1-\mu }{\sigma }------\left(1\right)$

  $2.05=\frac{2-\mu }{\sigma }------\left(2\right)$

Multiplying both sides of the equations by and subtracting yields

  $-2.3\sigma =-1$

or $\sigma =0.4348\text{minutes}$

Substituting this value back into equation (1), we obtain

  $-0.25=\frac{1-\mu }{0.4348}$

or $\mu =1+0.25\left(0.4348\right)$

  $=1.1087\text{minutes}$

Hence, the required mean and standard deviation are 0.4348min and 1.1087min respectively.

Conclusion:

Hence, the required mean and standard deviation are 0.4348min and 1.1087min respectively.