Question

Let x be a continuous random variable that has a normal distribution with a mean of 118.4 and a standard deviation of 19.3. Find to 4 decimal places the probability that x assumes a valuea. between 76.4 and 97.8.Probability = Enter you answer in accordance to the item a) of the question statementb. between 88.5 and 150.4.Probability = Enter you answer in accordance to the item b) of the question statement

Answer 1

Given that:

Mean = 118.4

Standard devation = 19.3

The z-score is

$Z=\frac{X-\mu}{\sigma}$

a. Find the probabaility between 76.4 and 97.8.

$\begin{gathered} P(76.4\leq X\leq97.8)=P(\frac{76.4-118.4}{19.3}\leq Z\leq\frac{97.8-118.4}{19.3}) \\ =P(-2.176\leq Z\leq-1.067) \\ =P(1.067\leq Z\leq2.176) \\ =P(0\leq Z\leq2.18)-P(0\leq Z\leq1.07) \end{gathered}$

From the standard normal table,

$\begin{gathered} P(76.4\leq X\leq97.8)=\text{0}.4854-0.3577 \\ =0.1277 \end{gathered}$

b. Find probability between 88.5 and 150.4.

$\begin{gathered} P(88.5\leq X\leq150.4)=P(\frac{88.5-118.4}{19.3}\leq Z\leq\frac{150.4-118.4}{19.3}) \\ =P(-1.55\leq Z\leq1.66) \\ =P(0\leq Z\leq1.55)+P(0\leq Z\leq1.66) \end{gathered}$

From the standard normal table,

$\begin{gathered} P(88.5\leq X\leq150.4)=\text{0}.4394+0.4515 \\ =0.8855 \end{gathered}$