Question

the amounts of time per workout an athlete uses a stairclimber are normally distributed, with a mean of 22 minutes and a standard deviation of 7 minutes. find the probability that a randomly selected athlete uses a stairclimber for (a) less than 18 minutes, (b) between 22 and 29 minutes, and (c) more than 30 minutes.
(a) the probability that a randomly selected athlete uses a stairclimber for less than 18 minutes is
(round to four decimal places as needed.)

Explanation:

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

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 18$, $\mu=22$, and $\sigma = 7$. So $z=\frac{18 - 22}{7}=\frac{-4}{7}\approx - 0.5714$.

Step2: Find the probability using the standard normal table

We want $P(X\lt18)$, which is equivalent to $P(Z\lt - 0.5714)$. Looking up the value in the standard - normal table, $P(Z\lt - 0.5714)\approx0.2843$.

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

For $x_1 = 22$, $z_1=\frac{22 - 22}{7}=0$. For $x_2 = 29$, $z_2=\frac{29 - 22}{7}=1$.

Step4: Find the probability for part (b)

We want $P(22\lt X\lt29)$, which is $P(0\lt Z\lt1)$. Since $P(0\lt Z\lt1)=P(Z\lt1)-P(Z\lt0)$. From the standard - normal table, $P(Z\lt1) = 0.8413$ and $P(Z\lt0)=0.5$, so $P(0\lt Z\lt1)=0.8413 - 0.5=0.3413$.

Step5: Calculate the z - score for part (c)

For $x = 30$, $z=\frac{30 - 22}{7}=\frac{8}{7}\approx1.1429$.

Step6: Find the probability for part (c)

We want $P(X\gt30)$, which is equivalent to $P(Z\gt1.1429)$. Since $P(Z\gt1.1429)=1 - P(Z\lt1.1429)$. From the standard - normal table, $P(Z\lt1.1429)\approx0.8735$, so $P(Z\gt1.1429)=1 - 0.8735 = 0.1265$.

Answer:

(a) $0.2843$
(b) $0.3413$
(c) $0.1265$