Question

suppose a normal distribution has a mean of 26 and a standard deviation of 4. what is the probability that a data value is between 28 and 31? round your answer to the nearest tenth of a percent.

a. 19.3%

b. 20.3%

c. 22.3%

d. 21.3%

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the data value.
For $x = 28$:
$z_1=\frac{28 - 26}{4}=\frac{2}{4}=0.5$
For $x = 31$:
$z_2=\frac{31 - 26}{4}=\frac{5}{4}=1.25$

Step2: Find the probabilities for z - scores

We use the standard normal distribution table (or z - table) to find the cumulative probabilities for $z_1 = 0.5$ and $z_2=1.25$.
From the z - table, $P(Z\lt0.5)=0.6915$ and $P(Z\lt1.25) = 0.8944$.

Step3: Calculate the probability between the two z - scores

The probability that a data value is between 28 and 31 is $P(0.5\lt Z\lt1.25)=P(Z\lt1.25)-P(Z\lt0.5)$
$P(0.5\lt Z\lt1.25)=0.8944 - 0.6915=0.2029$
Convert this to a percentage: $0.2029\times100 = 20.29\%\approx20.3\%$

Answer:

B. 20.3%