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)

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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