Area under a normal curve using Excel
Area Under a Normal Curve Using Excel
Norm Dot Dist Command
- Used to compute the area under a normal curve in Excel.
- Takes four arguments:
- x: The value from the population for which we compute the area to the left.
- Mean: The mean of the population.
- Standard Deviation: The standard deviation of the population.
- True/False: A true/false value; always use true for cumulative probability.
Example 1: Pregnancy Length
- Pregnancy length is approximately normally distributed.
- Mean (\mu) = 272 days
- Standard deviation (\sigma) = 9 days
Proportion of Pregnancies Lasting Less Than 265 Days
- Use
NORM.DIST command. - Arguments: x = 265, mean = 272, standard deviation = 9, cumulative = TRUE.
- Excel command:
=NORM.DIST(265, 272, 9, TRUE) - Result: 0.21835. This means approximately 21.84% pregnancies last less than 265 days.
Proportion of Pregnancies Lasting Longer Than 280 Days
- Find the area to the left of x = 280 using
NORM.DIST. - Subtract this area from 1 to find the area to the right.
- Excel command:
=1 - NORM.DIST(280, 272, 9, TRUE) - Result: 0.1870. This means approximately 18.7% pregnancies last longer than 280 days.
Proportion of Full-Term Pregnancies (252 to 298 Days)
- Find the area to the left of 252 and 298 using
NORM.DIST. - Subtract the smaller area from the larger area to find the area between.
- Excel command:
=NORM.DIST(298, 272, 9, TRUE) - NORM.DIST(252, 272, 9, TRUE) - Result: 0.9849. This means approximately 98.49% of pregnancies are full term.
Finding a Normal Value Corresponding to a Given Area
Norm Dot Inverse Command
- Used to find the value from a normal population with a given area to its left.
- Takes three arguments:
- Area (to the left)
- Mean
- Standard Deviation
- The mean has an area of 0.5 to both its right and left.
Example 2: IQ Scores
- IQ scores are normally distributed.
- Mean (\mu) = 100
- Standard deviation (\sigma) = 15
IQ Score Separating the Upper 2% from the Rest
- Find the score x_1 such that the area to its right is 0.02.
- The area to the left of x_1 is 0.98 (1 - 0.02 = 0.98).
- Use
NORM.INV command. - Arguments: area = 0.98, mean = 100, standard deviation = 15.
- Excel command:
=NORM.INV(0.98, 100, 15) - Result: 130.806. Rounded to the nearest whole number equals 131.
IQ Scores Separating the Middle 90% from the Rest
- Find scores x1 and x2 that bound the middle area of 0.9.
- The area of the two tails combined is 0.1.
- The area of each tail is 0.05 (0.1 / 2 = 0.05).
- The area to the left of x1 is 0.05, and to the left of x2 is 0.95.
- For x_1:
NORM.INV(0.05, 100, 15) = 75.3272 - For x_2:
NORM.INV(0.95, 100, 15) = 124.67287 - Rounded to whole numbers, 90% of IQ scores are between 75 and 125.
Example 3: Smartphone Apps
- Number of apps on a smartphone is normally distributed.
- Mean (\mu) = 90
- Standard deviation (\sigma) = 25
Finding the Third Quartile
- The third quartile is the 75th percentile (0.75).
- Use
NORM.INV command. - Arguments: area = 0.75, mean = 90, standard deviation = 25.
- Excel command:
=NORM.INV(0.75, 90, 25) - Result: 106.86
Using Table A2
- Find the area closest to 0.75 in the body of the table.
- The corresponding z-score is approximately 0.67.
- Compute x_1 = \mu + z \times \sigma
- x_1 = 90 + 0.67 \times 25 = 106.75
- The third quartile is approximately 106.75.