Consider a normal distribution with mean 30 and standard deviation 6. What is the probability a value selected at random from this distribution is greater than 30? (Round your answer to two decimal places.)
In a normal distribution, mean = median (50th percentile). If 50% ≤ 30, what percent are more?
To find the probability that a value selected at random from a normal distribution is greater than 30, we need to calculate the area under the curve to the right of 30.
First, we need to standardize the value of 30 using the z-score formula:
z = (x - mean) / standard deviation
Let's plug in the values:
z = (30 - 30) / 6 = 0 / 6 = 0
The z-score for a value of 30 is 0.
Now, we can look up the probability associated with this z-score in a standard normal distribution table or use a calculator.
Using a standard normal distribution table, we find that the probability of a value being greater than 0 (corresponding to a z-score of 0) is approximately 0.5000.
Therefore, the probability that a value selected at random from the given normal distribution is greater than 30 is approximately 0.50 or 50%.