Last year, the personal best high jumps of track athletes in a nearby state were normally distributed with a mean of 229 cm and a standard deviation of 19 cm. What is the probability that a randomly selected high jumper has a person best between 229 and 248 cm?
How would I solve this problem?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
I have an Algebra 2 book
To solve this problem, we can use the Z-score formula to standardize the values and then look up the corresponding probabilities in the standard normal distribution table.
The formula for calculating the Z-score is:
Z = (X - μ) / σ
where:
- Z is the standard score
- X is the value we want to standardize
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, we want to find the probability that a randomly selected high jumper has a personal best between 229 and 248 cm. We will calculate the Z-scores for both values and then find the cumulative probabilities.
For 229 cm:
Z1 = (229 - 229) / 19 = 0
For 248 cm:
Z2 = (248 - 229) / 19 ≈ 1
Once we have the Z-scores, we can use the standard normal distribution table to find the cumulative probabilities. The cumulative probability is the area under the curve from negative infinity up to the Z-score.
Using the standard normal distribution table, we can find the cumulative probabilities:
- The cumulative probability for Z1 = 0 is 0.5000
- The cumulative probability for Z2 ≈ 1 is 0.8413
To find the probability between these two values, we subtract the cumulative probability for Z1 from the cumulative probability for Z2.
Probability = P(0 < Z < 1) = P(Z < 1) - P(Z < 0)
Probability = 0.8413 - 0.5000
Probability = 0.3413
Therefore, the probability that a randomly selected high jumper has a personal best between 229 and 248 cm is approximately 0.3413, or 34.13%.