The mean time for running a mile is 450 se4conds with a standard deviation of 40 seconds. What time would you need to beat to be in the top 10%?
Assuming normal distribution, the z-score for 90% is 1.288, which means that the time is μ-1.288σ
=450-40*1.288 seconds.
For z-score, see:
http://www.regentsprep.org/Regents/math/algtrig/ATS7/ZChart.htm
6.6413
To find the time you need to beat to be in the top 10%, we first need to find the z-score corresponding to the top 10% of the distribution. The z-score represents the number of standard deviations a particular value is away from the mean.
To find the z-score, we can use the formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value we want to find the z-score for
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, we want to find the z-score for the top 10% of the distribution, which corresponds to a percentile of 90%. Using a standard normal distribution table or a calculator, we can find the z-score that corresponds to a percentile of 90%, which is approximately 1.28.
Now, we can rearrange the formula for the z-score to solve for x:
x = μ + (z * σ)
Plugging in the values, we get:
x = 450 + (1.28 * 40)
Calculating this, we find:
x ≈ 450 + 51.2
Therefore, to be in the top 10%, you would need to beat a time of approximately 501.2 seconds.