A normal distribution has a mean of µ = 100 with standard deviation = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 80 and X = 100?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion between the two Z scores.
so confused
To calculate the probability that a randomly selected score will have a value between X = 80 and X = 100 in a normal distribution with a mean of µ = 100 and standard deviation = 20, we can use the Z-score formula and the standard normal distribution table.
The Z-score formula is: Z = (X - µ) / σ, where X is the observed value, µ is the mean, and σ is the standard deviation.
First, we need to calculate the Z-score for X = 80. Using the formula, we substitute the values: Z1 = (80 - 100) / 20 = -1.
Next, we calculate the Z-score for X = 100: Z2 = (100 - 100) / 20 = 0.
Now, we look up the corresponding cumulative probability for the Z-scores in the standard normal distribution table. The cumulative probability represents the area under the curve to the left of the Z-score.
From the table, we find the cumulative probability for Z1 = -1 is approximately 0.1587, and for Z2 = 0 is 0.5.
To find the probability between X = 80 and X = 100, we subtract the cumulative probability for Z1 from the cumulative probability for Z2: P(80 < X < 100) = P(Z1 < Z < Z2) = P( -1 < Z < 0) = P(Z < 0) - P(Z < -1) = 0.5 - 0.1587 = 0.3413.
Therefore, the probability that a randomly selected score will have a value between X = 80 and X = 100 is approximately 0.3413, or 34.13%.