91影视

Ricardo brushing his teeth about $=40\%$of the time.

He spends more than$=2\%$of the time.

Using the standard normal table, $p$(brushing teeth for less than one minute) is equal to $0.40$, and the $z$value corresponds to $-0.2534.$

Additionally, $p$(brushing for over two minutes) $=0.02$.

Resulting in brushing teeth for less than two minutes being $98\%=0.98$.

The standard normal table gives a $z$value of $2.054$for a probability of $0.98$.

The following graph illustrates how to set up two equations with two unknowns based on the information given.

Following these two equations, we can calculate the unknown mean and standard deviation.

localid="1649998428756" $$\begin{gathered} -0.25=\frac{1-渭}{蟽}---1 \\ 2.05=\frac{2-渭}{蟽}---2 \end{gathered}$$

By multiplying by and subtracting from both equations, we get

$$\begin{gathered} -2.3蟽=-1 \\ or蟽=0.4343minutes \end{gathered}$$

By replacing this value in equation 1 , we obtain

localid="1649998443023" $$\begin{gathered} 0.25=\frac{1-渭}{0.4348} \\ or渭=1+0.25(0.4348) \\ =1.1087minutes \end{gathered}$$

As a result, the required mean and standard deviation are$0.4348minand1.1087min$respectively.