On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percentage of scores fall between 70 and 80?

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N()

The z probability is given by;

           Z = ~ N(0,1)    where, = 75  and   = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( < ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( < ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

                                              = 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)