Suppose over several years of offering AP Statistics, a high school finds that final exam scores are normally distributed with a mean of 78 and a standard deviation of 6.a. What are the mean, standard deviation, and shape of the distribution of x-bar for n = 50?b. What’s the probability a sample of scores will have a mean greater than 80?c. Sketch the distribution curve for part B, showing the area that represents the probability you found.

ux = 78

s= 0.8485

P ( z>0.33)  = 0.3707

Step-by-step explanation:

According to central limit theorem the mean of the population is equal to the mean of the sample and the standard deviation is found by dividing the standard deviation of the population with the square root of the sample size

Thus

ux = u= 78

σ= 6

s= σ/√n= 6/√50= 6/7.07106= 0.8485

We find the z score

Z= X- u / σ

z= 80-78/6

z= 2/6= 1/3=0.333

P ( z>0.33) = 1 - P(z=0.33)

                  = 1- 0.6293

                     = 0.3707

We see in the figure that the blue area represents the area greater than z= 0.333 or area greater than 1/3.

The probability of z= 0.33 from the table is 0.6293

We find the probability of area greater than 0.33 by subtracting from 1 .

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: