use the normal distribution to the right to answer the questions. (a) what percent of the scores are less…

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.)

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.)

Answer

Answer:

(a) 36.32% (b) 735

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Here, $x = 19$, $\mu=20.9$, and $\sigma = 5.4$. So, $z=\frac{19 - 20.9}{5.4}=\frac{-1.9}{5.4}\approx - 0.35$.

Step2: Find the percentile for part (a)

Using a standard normal distribution table or calculator, the area to the left of $z=-0.35$ is approximately $0.3632$ or $36.32%$.

Step3: Calculate z - score for part (b)

For $x = 21$, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $z=\frac{21 - 20.9}{5.4}=\frac{0.1}{5.4}\approx0.02$.

Step4: Find the area to the right for part (b)

The area to the left of $z = 0.02$ is approximately $0.5080$ from the standard - normal table. The area to the right is $1 - 0.5080=0.4920$.

Step5: Calculate the number of scores for part (b)

Out of $n = 1500$ scores, the number of scores greater than 21 is $n\times0.4920=1500\times0.4920 = 738\approx735$ (after rounding).