Question

the results of a certain medical test are normally distributed with a mean of 123 and a standard deviation of 15. 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 150 and 162.
click the icon to view the table of z-scores and percentiles.
the percentage of people with readings between 150 and 162 is
(round to two decimal places as needed.)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 123\) (mean) and \(\sigma=15\) (standard deviation).
For \(x = 150\):
\(z_1=\frac{150 - 123}{15}=\frac{27}{15}=1.8\)
For \(x = 162\):
\(z_2=\frac{162-123}{15}=\frac{39}{15}=2.6\)

Step2: Find the corresponding percentiles

Using the standard normal table (z - score table):
The percentile corresponding to \(z = 1.8\) is \(P(Z<1.8)=0.9641\)
The percentile corresponding to \(z = 2.6\) is \(P(Z < 2.6)=0.9953\)

Step3: Calculate the percentage between the two z - scores

The percentage of people with readings between \(150\) and \(162\) is \(P(1.8<Z<2.6)=P(Z < 2.6)-P(Z<1.8)\)
\(P(1.8 < Z<2.6)=0.9953 - 0.9641=0.0312\)

Answer:

\(3.12\%\)