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 450 seconds and a standard deviation of 50 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 335 seconds.
Is the answer 0.0107? Anyone please
can some one please help me!!!
You are given the normal model N(450, 50)
Find the z score for 335 seconds.
z = (x-u)/s
where u = mu and s = sigma.
z = (335-450)/50
z = -2.3
Now use the normal cumulative distribution function on a calculator or program to find the P(z<-2.3)
normalCDF(-99,-2.3) = .0107
Marth this are my choices
A) 0.9893
B) 0.0107
C) 0.4893
D) 0.5107
So I'm assuming it's letter B? correct
Dahhhh lol
To find the probability that a randomly selected boy in secondary school can run the mile in less than 335 seconds, we need to calculate the z-score and then use a standard normal distribution table.
The z-score represents how many standard deviations a value is from the mean. It can be calculated using the formula:
z = (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.
In this case, we have:
x = 335 seconds
μ = 450 seconds
σ = 50 seconds
Substituting these values into the equation gives:
z = (335 - 450) / 50
z = -2.3
Now we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score.
Looking up the z-score of -2.3 in the table or using a calculator, we find that the probability is approximately 0.0107.
Therefore, the probability that a randomly selected boy in secondary school can run the mile in less than 335 seconds is indeed approximately 0.0107.