Final Answer:
Approximately 10.92% of a standard normal distribution N(μ=0,σ=1) is found in the region z>-1.13.
Step-by-step explanation:
A normal distribution is defined by two parameters, the mean (μ) and the standard deviation (σ). A standard normal distribution has a mean of 0 and a standard deviation of 1. The percentage of the standard normal distribution that is found in each region can be determined by first calculating the area under the curve for the given region. The area under the curve for the region z>-1.13 can be determined by calculating the area of the two tails on either side of the region.
The area of the two tails is equal to the area of the entire normal distribution, which is 1. The area of the left tail is equal to the area of the right tail and can be calculated using the cumulative distribution function. For a normal distribution, the cumulative distribution function is given by the equation F(x) = P(X<=x).
The cumulative distribution function for the left tail can be calculated by substituting the z-score -1.13 into the equation. This yields F(-1.13) = 0.4. The area of the left tail is equal to 1-F(-1.13), which is 0.6. The area of the right tail is the same as the area of the left tail and can be calculated using the same equation, yielding F(1.13) = 0.9. The area of the right tail is equal to 1-F(1.13), which is 0.1. The total area of the two tails is equal to the sum of the areas of the left and right tails, which is 0.6+0.1=0.7. The percentage of the standard normal distribution that is found in the region z>-1.13 is equal to 0.7/1, which is approximately 10.92%.