Sketch a standard normal curve and shade the area that lies to the right of [tex]$a .-1.23, b .0 .5, c .0$[/tex], and [tex]$d .4 .2$[/tex]. Then determine the area under the standard normal.
The area to the right of 0.5 is $\square$.
(Round to four decimal places as needed.)
Answer (2)
Find the area to the left of 0.5 using the standard normal CDF: P ( Z < 0.5 ) ≈ 0.6915 .
Calculate the area to the right of 0.5 by subtracting the area to the left from 1: 0.5) = 1 - P(Z < 0.5)"> P ( Z > 0.5 ) = 1 − P ( Z < 0.5 ) .
Perform the subtraction: 1 − 0.6915 = 0.3085 .
The area to the right of 0.5 is: 0.3085 .
Explanation
Understand the problem and provided data We are asked to find the area under the standard normal curve to the right of 0.5. The standard normal curve has a mean of 0 and a standard deviation of 1.
Define the area to calculate To find the area to the right of 0.5, we need to calculate 0.5)"> P ( Z > 0.5 ) , where Z is a standard normal random variable. This represents the probability that a standard normal random variable is greater than 0.5.
Use CDF to find the area We can use the standard normal cumulative distribution function (CDF) to find the area to the left of 0.5, which is P ( Z < 0.5 ) . Then, we can subtract this value from 1 to find the area to the right of 0.5, since the total area under the standard normal curve is 1. That is, 0.5) = 1 - P(Z < 0.5)"> P ( Z > 0.5 ) = 1 − P ( Z < 0.5 ) .
Calculate the area to the right of 0.5 Using a standard normal table or a calculator, we find that P ( Z < 0.5 ) ≈ 0.6915 . Therefore, the area to the right of 0.5 is 0.5) = 1 - 0.6915 = 0.3085"> P ( Z > 0.5 ) = 1 − 0.6915 = 0.3085 .
State the final answer The area to the right of 0.5 under the standard normal curve, rounded to four decimal places, is 0.3085.
Examples
Consider a standardized test where the scores are normally distributed with a mean of 0 and a standard deviation of 1. If you want to know the percentage of students who scored above 0.5 standard deviations from the mean, you would calculate the area to the right of 0.5 under the standard normal curve. This area represents the proportion of students who performed better than average by a certain margin.
The area to the right of 0.5 under the standard normal curve is 0.3085 when rounded to four decimal places. This is calculated by first finding the area to the left of 0.5 and subtracting it from 1. The approximate area to the left of 0.5 is 0.6915.
;
