Final answer:
The probability that an individual distance is greater than 218.40 cm is approximately 0.0416. The probability that the mean for 15 randomly selected distances is greater than 202.80 cm is approximately 0.8661. The normal distribution can be used in part (b) even though the sample size does not exceed 30 due to the Central Limit Theorem.
Step-by-step explanation:
In order to solve this problem, we need to use the properties of a normal distribution. Let's solve each part of the question:
Part A:
To find the probability that an individual distance is greater than 218.40 cm, we need to calculate the z-score and then find the area under the normal distribution curve to the right of that z-score.
The formula to calculate the z-score is: z = (x - mean) / standard deviation.
Plugging in the given values, we get: z = (218.40 - 205) / 7.8 = 1.731.
Using a z-table or calculator, we can find that the area to the right of this z-score is approximately 0.0416.
Part B:
To find the probability that the mean for 15 randomly selected distances is greater than 202.80 cm, we need to calculate the standard error of the mean and then find the area under the normal distribution curve to the right of that mean.
The formula to calculate the standard error of the mean is: standard deviation / sqrt(sample size).
Plugging in the given values, we get: standard error = 7.8 / sqrt(15) = 2.014.
Then, we calculate the z-score using the formula: z = (x - mean) / standard error.
Plugging in the given values, we get: z = (202.80 - 205) / 2.014 = -1.113.
Using a z-table or calculator, we can find that the area to the right of this z-score is approximately 0.8661.
Part C:
The normal distribution can be used in part (b) even though the sample size does not exceed 30 because of the Central Limit Theorem. According to the Central Limit Theorem, as the sample size increases, the sampling distribution of the mean approaches a normal distribution regardless of the shape of the original population.