In 2002 John Daly led the PGA tour in driving distance with an average drive of 305.5 yards and a standard deviation of 17.2 yards. Assuming that these distances are normally distributed, 80% of John Daly's drives are longer than what distance?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion. Insert values in the equation above and solve for the score.
The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 170 and a variance of 4. The material is considered defective if the breaking strength is less than 166 pounds. What is the probability that a single, randomly selected piece of material will be defective?
To find the distance that 80% of John Daly's drives are longer than, we need to use the concept of z-scores and the standard normal distribution.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, the mean (μ) is 305.5 yards and the standard deviation (σ) is 17.2 yards. We want to find the distance (x) such that 80% of the drives are longer than that distance.
To find the z-score that corresponds to the 80th percentile, we need to find the z-score at which 80% of the area lies to the left of it. Using a standard normal distribution table (also known as a z-table) or a statistical calculator, we can look up the z-score that corresponds to an 80% cumulative probability.
Let's denote this z-score as "z_80".
Next, we rearrange the z-score formula to solve for x:
x = z * σ + μ
where z is the z-score we found from the previous step, σ is the standard deviation, and μ is the mean.
By substituting the values into the formula, we can calculate the distance (x) that corresponds to the desired percentile.
Therefore, to find the distance (x) such that 80% of John Daly's drives are longer than x, you need to find the z-score corresponding to the 80th percentile and then use that z-score to calculate the distance.