A distribution of values is normal with a mean of 225.5 and a standard deviation of 3.2.
Find the probability that a randomly selected value is greater than 222.6.
P(X > 222.6) =
Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
To find the probability that a randomly selected value is greater than 222.6, we need to use the standard normal distribution.
First, let's convert the value 222.6 to a z-score using the formula:
z = (x - mean) / standard deviation
z = (222.6 - 225.5) / 3.2
z = -0.9 / 3.2
z ≈ -0.2813
Next, we can find the probability using the standard normal distribution table (also known as the Z-table) or a calculator.
The Z-table provides the cumulative probability up to a given z-score. Since we want to find the probability that a value is greater than 222.6, we need to find the area to the right of the z-score -0.2813.
Looking up -0.2813 in the Z-table, we find that the cumulative probability is 0.3894.
However, this is the probability for values less than or equal to 222.6, so we subtract this value from 1 to find the probability of values greater than 222.6.
P(X > 222.6) = 1 - 0.3894
P(X > 222.6) ≈ 0.6106
Therefore, the probability that a randomly selected value is greater than 222.6 is approximately 0.6106.