Question

in the country of united states of heightlandia, the height measurements of ten - year - old children are approximately normally distributed with a mean of 54.7 inches, and standard deviation of 5.3 inches. a) what is the probability that a randomly chosen child has a height of less than 53.25 inches? answer= (round your answer to 3 decimal places.) enter an integer or decimal number more... b) what is the probability that a randomly chosen child has a height of more than 63.4 inches? answer= (round your answer to 3 decimal places.)

Explanation:

Step1: Calculate the z - score for part A

Use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $x = 53.25$, $\mu=54.7$, $\sigma = 5.3$.
$z_A=\frac{53.25 - 54.7}{5.3}=\frac{-1.45}{5.3}\approx - 0.274$

Step2: Find the probability for part A

Look up the z - score in the standard normal distribution table. $P(X < 53.25)=P(Z < - 0.274)$. From the standard - normal table, $P(Z < - 0.274)\approx0.392$

Step3: Calculate the z - score for part B

Use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $x = 63.4$, $\mu = 54.7$, $\sigma=5.3$.
$z_B=\frac{63.4 - 54.7}{5.3}=\frac{8.7}{5.3}\approx1.642$

Step4: Find the probability for part B

$P(X>63.4)=1 - P(X\leqslant63.4)=1 - P(Z\leqslant1.642)$. From the standard - normal table, $P(Z\leqslant1.642)\approx0.949$, so $P(X > 63.4)=1 - 0.949 = 0.051$

Answer:

A) 0.392
B) 0.051