Normal distribution HELP Suppose that is normally distributed with mean 80 and standard deviation 13.

What is the probability that is greater than 103.53?
What value of does only the top 16% exceed?

0.0351478382 (To be precise)

Step-by-step explanation: Can I get brainliest? Thanks

1. Normal Distribution --> Z ~ (0,1^2)

2. Use normalcdf(lower bound, upper bound, μ, σ) function on a graphing calculator

P(Z≥103.53) = normalcdf(103.53, 1e99 [default], 80, 13)

P(Z≥103.53) ≈ 0.03

3. μ+σ ≈ 13.59% According to Z-distribution chart

80+13=93

So about 93 exceed only the top 16% (estimated answer not exact)

the slope of the line is 60

one point on the line is (40,10)

step-by-step explanation:

in the equation outside the parenthesis the equation always tells you the slope. in this case it is 60.

in the parenthesis is also says 40 which is x because it is near the x and 10 is y because it is near y.