Question

the results of a certain medical test are normally distributed with a mean of 126 and a standard deviation of 19. 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 111 and 141. click the icon to view the table of z - scores and percentiles. the percentage of people with readings between 111 and 141 is % (round to two decimal places as needed)

Explanation:

Step1: Calculate the z - score for \(x = 111\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 126\) (mean) and \(\sigma=19\) (standard deviation).
For \(x = 111\), \(z_1=\frac{111 - 126}{19}=\frac{- 15}{19}\approx - 0.79\)

Step2: Calculate the z - score for \(x = 141\)

Using the same formula \(z=\frac{x-\mu}{\sigma}\)
For \(x = 141\), \(z_2=\frac{141-126}{19}=\frac{15}{19}\approx0.79\)

Step3: Find the percentage using the z - score table

The percentage of data to the left of \(z=-0.79\) is \(P(Z < - 0.79)=0.2148\)
The percentage of data to the left of \(z = 0.79\) is \(P(Z < 0.79)=0.7852\)
The percentage of data between \(z=-0.79\) and \(z = 0.79\) is \(P(-0.79<Z<0.79)=P(Z < 0.79)-P(Z < - 0.79)\)
\(P(-0.79 < Z < 0.79)=0.7852-0.2148 = 0.5704\)

Answer:

\(57.04\%\)