A normal distribution has a mean of 20 and a standard deviation of 3.What is the probability that a randomly selected x-value from the distribution is: A)at least 14?

B)at most 26?
C)Between 8 and 23?

To find the probability for each case, we need to use the properties of the normal distribution and standardize the values using a Z-score.

The Z-score is calculated by subtracting the mean from the given value and dividing it by the standard deviation:

Z = (x - μ) / σ

where x is the given value, μ is the mean, and σ is the standard deviation.

A) To find the probability that x is at least 14, we need to calculate the area to the right of the value 14.

Z = (14 - 20) / 3 = -2

Using a standard normal distribution table or a calculator, we can find that the area to the right of Z = -2 is approximately 0.9772. However, we need to find the area to the left, so we subtract this value from 1:

P(x ≥ 14) = 1 - 0.9772 = 0.0228 or 2.28%

B) To find the probability that x is at most 26, we need to calculate the area to the left of the value 26.

Z = (26 - 20) / 3 = 2

Using the standard normal distribution table or a calculator, we can find that the area to the left of Z = 2 is approximately 0.9772.

P(x ≤ 26) = 0.9772 or 97.72%

C) To find the probability that x is between 8 and 23, we need to calculate the area between the two values.

First, find the Z-score for each value:
Z1 = (8 - 20) / 3 = -4
Z2 = (23 - 20) / 3 = 1

Next, find the area to the left of each Z-score using the standard normal distribution table or a calculator:
Area1 = 0.0002 (approximately)
Area2 = 0.8413 (approximately)

Since we want the probability between the two values, we subtract the smaller area from the larger area:

P(8 ≤ x ≤ 23) = Area2 - Area1 = 0.8413 - 0.0002 = 0.8411 or 84.11%

So, the probabilities for the given x-values from the distribution are:
A) P(x ≥ 14) = 0.0228 or 2.28%
B) P(x ≤ 26) = 0.9772 or 97.72%
C) P(8 ≤ x ≤ 23) = 0.8411 or 84.11%