The combined (verbal + quantitative reasoning) score on the GRE (Graduate Record Exam) is normally distributed with mean 1049 and standard deviation 189. Suppose n = 16 randomly selected students take the GRE on the same day. a. What is the probability that one randomly selected student scores above 1100 on the GRE?
b. Describe the sampling distribution of the sample mean for the 16 students (description should include mean, standard deviation and shape).
c. What is the probability that a random sample of 16 students has a mean GRE score that is less than 1100?
d. What is the probability that a random sample of 16 students has a mean score that is more than 1100?
e. If the sample size in (d) above was increased to 64, what affect would it have on the probability (more, less, the same, can't tell) and why?

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Step-by-step explanation:

Given the following :

Normal distribution :

Mean (m) = 1049

Standard deviation (sd) = 189

Number of samples (n) = 16

A) What is the probability that one randomly selected student scores above 1100 on the GRE?

Obtain the z score

Zscore = (x - m) / sd

Zscore = (1100 - 1049) / 189

Zscore = 51 / 189 = 0.2698412 = 0.27

Using the z-table

P(X > 100) = 1 - P(X < 100) = 1 - 0.6064 = 0.3936

B) The sample mean = population mean

Hence sample mean = 1049

The sample standard deviation = standard error

Standard Error (SE) = sd / sqrt(n)

SE = 189/4 = 47.25

Hence shape = normal, standard deviation = 47.25, mean = 1049

C.) P(X < 1100)

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 47.25

Zscore = 51 / 47.25 = 1.0793650 = 1.08

Using the z-table :

P(X < 1100) = 0.8599

D.) P(X > 1100)

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 47.25

Zscore = 51 / 47.25 = 1.0793650 = 1.08

Using the z-table :

P(X > 1100) = 1 - P(X < 1100) = (1 - 0.8599) = 0.1401

E) if n = 64

SE = 189 / sqrt(64) = 189 / 8 = 23.625

P(X > 1100)

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 23.625

Zscore = 51 / 47.25 = 2.1587301 = 2.16

Using the z-table :

P(X > 1100) = 1 - P(X < 1100) = (1 - 0.9846) = 0.0154

The probability decreases as the same distribution hinges towards a Boal distribution with increasing sample size.

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

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