the times of the runners in a marathon are normally distributed, with a mean of 3 hours and 50 minutes and a…
the times of the runners in a marathon are normally distributed, with a mean of 3 hours and 50 minutes and a standard deviation of 30 minutes. what is the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes? use the portion of the standard normal table below to help answer the question.\n\n| z | probability |\n|----|----| \n| 0.00 | 0.5000 |\n| 0.50 | 0.6915 |\n| 1.00 | 0.8413 |\n| 2.00 | 0.9772 |\n| 3.00 | 0.9987 |\n\n16%\n32%\n34%\n84%
Answer
Explanation:
Step1: Convert times to minutes
3 hours and 50 minutes = 3×60 + 50 = 230 minutes. 3 hours and 20 minutes = 3×60+20 = 200 minutes.
Step2: Calculate the z - score
The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 230$, $\sigma=30$, and $x = 200$. So $z=\frac{200 - 230}{30}=\frac{- 30}{30}=-1$.
Step3: Use the symmetry of the normal distribution
The standard - normal table gives probabilities for $z\geq0$. Since the normal distribution is symmetric about $z = 0$, the probability $P(Z\leq - 1)=P(Z\geq1)$. And $P(Z\leq1)=0.8413$, so $P(Z\leq - 1)=1 - 0.8413 = 0.1587\approx16%$.
Answer:
16%