A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 440 seconds and a standard deviation of 60 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 302 seconds.
A) 0.5107
B) 0.0107
C) 0.9893
D) 0.4893
Z = (score-mean)?SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Z = (score-mean)/SD
.0107
To find the probability that a randomly selected boy in secondary school can run the mile in less than 302 seconds, we need to calculate the area under the normal distribution curve up to the z-score corresponding to 302 seconds.
To do this, we first need to standardize the value of 302 seconds using the formula for z-score:
z = (x - μ) / σ
where x is the observed value, μ is the mean, and σ is the standard deviation.
In this case, x = 302 seconds, μ = 440 seconds, and σ = 60 seconds.
Substituting the values into the formula, we get:
z = (302 - 440) / 60
z ≈ -2.3
Now, we find the corresponding probability using a standard normal distribution table or a calculator. The probability associated with a z-score of -2.3 is approximately 0.0107.
Therefore, the correct answer is B) 0.0107.