Question

Assume that $x$ has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) $\mu =29;\sigma =3.4P(x\ge 30)=$___.

Answer 1

Step-1: Find the $z$-score for the present situation:

In a normally distributed random variable $x$ with mean $\mu$ and standard deviation $\sigma$, the probability $P(z<c)$ that the $z$-score $z=\frac{x-\mu }{\sigma }$ is less than a particular number $c$ can be looked up from a standard normal distribution $z$-score table.

It is given that normally distributed random variable $x$ has mean $\mu =29$ and standard deviation $\sigma =3.4$. It is required to find $P(x\ge 30)$.

Find the $z$-score $z=\frac{x-\mu }{\sigma }$ for $x=30$, $\mu =29$, and $\sigma =3.4$.

$\begin{array}{rcl}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{30-29}{3.4}\\ & =& \frac{1}{3.4}\\ & \approx & 0.29\end{array}$

Thus the probability $P(x\ge 30)$ is equal to $P(z\ge 0.29)$.

Step-2: Write the required probability in terms of the probability of its complement event:

Note that the event $z\ge 0.29$ is complement of the event $z<0.29$. Since the probabilities of complementary events add up to $1$, write $P(z\ge 0.29)=1-P(z<0.29)$.

Step-3: Calculate the required probability:

Look up $P(z<0.29)$ from a standard normal distribution $z$-score table to get $P(z<0.29)=0.61409$. Substitute this value in the equation $P(z\ge 0.29)=1-P(z<0.29)$.

$\begin{array}{rcl}P(z& \ge & 0.29)=1-P(z<0.29)\\ & =& 1-0.61409\\ & =& 0.38591\end{array}$

Round off to four decimal places and write $\mathit{P}\mathbf{(}\mathit{z}\mathbf{\ge }\mathbf{0}\mathbf{.}\mathbf{29}\mathbf{)}\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{3859}$.

Hence, $\begin{array}{rcl}\mathbf{P}\mathbf{(}\mathbf{z}& \mathbf{\ge }& \mathbf{0}\mathbf{.}\mathbf{29}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{38591}\end{array}$.