Question

use the normal distribution to the right to answer the questions. (a) what percent of the scores are less than 19? (b) out of 1500 randomly selected scores, about how many would be expected to be greater than 21? (a) the percent of scores that are less than 19 is % (round to two decimal places as needed.) standardized test composite scores μ = 21.2 σ = 5.5 19 21 score

Explanation:

Step1: Calculate the z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Given $\mu = 21.2$, $\sigma=5.5$ and $x = 19$. Then $z=\frac{19 - 21.2}{5.5}=\frac{-2.2}{5.5}=- 0.4$.

Step2: Find the proportion for the z - score

Using a standard normal distribution table or calculator, the proportion of values to the left of $z=-0.4$ is $P(Z < - 0.4)=0.3446$. To convert this to a percentage, we multiply by 100. So the percentage of scores less than 19 is $0.3446\times100 = 34.46\%$.

Step3: Calculate the z - score for part (b)

For $x = 21$, $z=\frac{21 - 21.2}{5.5}=\frac{-0.2}{5.5}\approx - 0.04$.

Step4: Find the proportion of scores greater than 21

The proportion of values to the right of $z=-0.04$ is $P(Z>-0.04)=1 - P(Z < - 0.04)$. From the standard - normal table, $P(Z < - 0.04)=0.4840$, so $P(Z>-0.04)=1 - 0.4840 = 0.5160$.

Step5: Calculate the number of scores greater than 21 out of 1500

We multiply the proportion by the total number of scores. So the number of scores greater than 21 out of 1500 is $0.5160\times1500 = 774$.

Answer:

(a) 34.46%
(b) 774