A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean 7.11 minutes and standard deviation 0.74 minute. Choose a student at random from this group and call his time for the mile Y (a) Write the event “the student could run a mile in less than 6.18 minutes” in terms of the value of the random variable Y Use the symbols “< or “> as appropriate to indicate the bounds on Y (b) What is the probability of the event from part (a)?
The Correct Answer and Explanation is:
a) Event in terms of the random variable Y:
The random variable YYY represents the time, in minutes, it takes for a randomly selected student to run a mile. The problem is asking for the event where a student’s mile time is less than 6.18 minutes. This event can be written in terms of the random variable YYY as:Y<6.18Y < 6.18Y<6.18
This means that we are interested in the probability that the time for a randomly selected student is less than 6.18 minutes.
(b) Probability of the event from part (a):
To calculate the probability, we need to use the fact that the distribution of the times is approximately Normal with a mean μ=7.11\mu = 7.11μ=7.11 minutes and a standard deviation σ=0.74\sigma = 0.74σ=0.74 minutes.
To find the probability that Y<6.18Y < 6.18Y<6.18, we first need to standardize the value using the Z-score formula:Z=Y−μσZ = \frac{Y – \mu}{\sigma}Z=σY−μ
Substitute the known values into this formula:Z=6.18−7.110.74=−0.930.74≈−1.26Z = \frac{6.18 – 7.11}{0.74} = \frac{-0.93}{0.74} \approx -1.26Z=0.746.18−7.11=0.74−0.93≈−1.26
Next, we need to find the probability that Z<−1.26Z < -1.26Z<−1.26. To do this, we can use the standard normal distribution table (or a calculator with the normal distribution function) to find the cumulative probability associated with a Z-score of -1.26.
Looking up Z=−1.26Z = -1.26Z=−1.26 in the standard normal table, we find that the cumulative probability is approximately 0.103.
Thus, the probability that the student could run a mile in less than 6.18 minutes is:P(Y<6.18)≈0.103P(Y < 6.18) \approx 0.103P(Y<6.18)≈0.103
Conclusion:
The probability that a randomly chosen student at the University of Illinois could run a mile in less than 6.18 minutes is approximately 0.103, or about 10.3%. This means there is about a 10.3% chance that a randomly selected student from this group would run the mile in under 6.18 minutes.
