Question

the results of a certain medical test are normally distributed with a mean of 123 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 118 and 128. click the icon to view the table of z - scores and percentiles. the percentage of people with readings between 118 and 128 is \\( \\% \\). (round to two decimal places as needed.)

Explanation:

Step1: Calculate z-score for 118

The formula for z-score is $z = \frac{x - \mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 123$, $\sigma = 19$, and $x = 118$.
So, $z_1 = \frac{118 - 123}{19} = \frac{-5}{19} \approx -0.2632$

Step2: Calculate z-score for 128

Using the same formula, with $x = 128$:
$z_2 = \frac{128 - 123}{19} = \frac{5}{19} \approx 0.2632$

Step3: Find the area between z-scores

We need to find $P(-0.2632 < Z < 0.2632)$. Using the standard normal table (or z-table), the area to the left of $z = 0.26$ is approximately 0.6026, and the area to the left of $z = -0.26$ is approximately 0.3974. Subtracting these gives the area between them: $0.6026 - 0.3974 = 0.2052$. Converting this to a percentage gives $0.2052 \times 100 = 20.52\%$ (rounded to two decimal places).

Answer:

20.52