Question

In a long jump competition modeled by a normal distribution with mean 5.2 m and variance 0.6 m^2 for a group of 30 competitors, how many would you expect to jump further than 6 m?

a) Approximately 12 competitors
b) Approximately 15 competitors
c) Approximately 18 competitors
d) Approximately 21 competitors

Answer 1

To find the number of competitors who would be expected to jump further than 6m in a long jump competition, we can use the properties of the normal distribution.

Step-by-step explanation:

To find the number of competitors who would be expected to jump further than 6m in a long jump competition, we can use the properties of the normal distribution. Given the mean and variance of the distribution, we can calculate the standard deviation by taking the square root of the variance. In this case, the standard deviation is √0.6m^2 = 0.77m.

Next, we can calculate the z-score for a jump of 6m, which tells us how many standard deviations away from the mean this distance is. The z-score formula is z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. In this case, the z-score for a jump of 6m is (6 - 5.2) / 0.77 = 1.04.

Finally, we can use a standard normal distribution table or a calculator to find the proportion of competitors who would jump further than 6m. According to the table, approximately 15.87% of competitors would jump further than 6m. Therefore, the answer is b) Approximately 15 competitors.