A season of football scores falls in a normal distribution with a mean of 35 and standard deviation of 6. What is the probability that a score was between 29 and 41?
z=(X-μ)/σ
Look up tables and find the difference between z(41) and z(29)
just enter you data into
http://davidmlane.com/hyperstat/z_table.html
you should get .682689
To solve this problem, we can use the standard normal distribution and the z-score formula. The z-score tells us how many standard deviations an individual score is from the mean.
First, we need to convert the given scores into z-scores. The formula for the z-score is:
z = (x - μ) / σ
Where:
- x is the given score
- μ is the mean of the distribution (35 in this case)
- σ is the standard deviation of the distribution (6 in this case)
For x = 29:
z1 = (29 - 35) / 6 = -1
For x = 41:
z2 = (41 - 35) / 6 = 1
Now, we need to find the probability that the z-score is between -1 and 1.
To find this probability, we can use a standard normal distribution table or a calculator that can provide the cumulative probability.
Using a standard normal distribution table, we find that the cumulative probability for z = -1 is approximately 0.1587, and the cumulative probability for z = 1 is approximately 0.8413.
To find the probability between -1 and 1, we subtract the cumulative probability for z = -1 from the cumulative probability for z = 1:
P(-1 < z < 1) = 0.8413 - 0.1587 = 0.6826
Therefore, the probability that a score falls between 29 and 41 is approximately 0.6826, or 68.26%.