A normal distribution has µ = 80 and standard deviation = 10. What is the probability of randomly selecting a score greater than 90 from this distribution
Find the z-score
90-80 divided by 10
Use the z-score to find the probability to the right of that score.
To find the probability of randomly selecting a score greater than 90 from a normal distribution, we can use the concept of standardization or z-scores.
Step 1: Calculate the z-score for the value 90 using the formula:
z = (x - µ) / σ
where x is the value, µ is the mean, and σ is the standard deviation.
z = (90 - 80) / 10
z = 10 / 10
z = 1
Step 2: Once we have the z-score, we can use a standard normal distribution table (also called the z-table) or a statistical calculator to find the area or probability associated with that z-score.
In this case, we want to find the probability of selecting a score greater than 90, which corresponds to the area under the curve to the right of the z-score of 1.
Using the z-table, we find that the area to the left of the z-score of 1 is approximately 0.8413. Therefore, the area to the right (or the probability of randomly selecting a score greater than 90) is 1 - 0.8413 = 0.1587.
So, the probability of randomly selecting a score greater than 90 from this normal distribution is approximately 0.1587, or 15.87%.