A normal distribution has a mean of µ = 70 with σ = 10. If one score is randomly selected from this distribution, what is the probability that the score will be greater than X = 82?

0.11507

Step-by-step explanation:

A normal distribution has a mean of µ = 70 with σ = 10. If one score is randomly selected from this distribution, what is the probability that the score will be greater than X = 82?

We solve this question using z score formula:

z = (x-μ)/σ,

where

x is the raw score = 82

μ is the population mean = 70

σ is the population standard deviation = 10

z = 82 - 70/10

z = 1.2

Probability value from Z-Table:

P(x<82) = 0.88493

P(x>82) = 1 - P(x<82) = 0.11507

Therefore, the probability that the score will be greater than X = 82 is

0.11507

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: