Question

the results of a certain medical test are normally distributed with a mean of 125 and a standard deviation of 10. convert the given results into z - scores, and then use the accompanying table of z - scores and percentiles to find the percentage of people with readings between 119 and 131.
click the icon to view the table of z - scores and percentiles.
the percentage of people with readings between 119 and 131 is %
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the z - score for \(x = 119\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 125\) (mean) and \(\sigma=10\) (standard deviation).
For \(x = 119\), \(z_1=\frac{119 - 125}{10}=\frac{-6}{10}=- 0.6\)

Step2: Calculate the z - score for \(x = 131\)

Using the same formula \(z=\frac{x-\mu}{\sigma}\)
For \(x = 131\), \(z_2=\frac{131 - 125}{10}=\frac{6}{10}=0.6\)

Step3: Find the percentiles corresponding to the z - scores

Looking up the z - scores in the standard normal table (z - score table):
The percentile corresponding to \(z=-0.6\) is \(27.43\%\)
The percentile corresponding to \(z = 0.6\) is \(72.57\%\)

Step4: Calculate the percentage between the two z - scores

The percentage of people with readings between \(119\) and \(131\) is \(P(119<X<131)=P(z_1<Z<z_2)\)
\(P(-0.6 < Z < 0.6)=72.57\% - 27.43\%=45.14\%\)

Answer:

\(45.14\)