C. Find the mean, variance and standard deviation of the following probability distributions. (5 pts.) X 1 6 11 16 21 P(X) 1/7 1/7 2/7 1/7 277 D. Given the population of 5,000 scores with p = 86 and o = 10. How many scores are; (10 pts.) A Between 96 to 1067 B. middle 50% of the distribution? E. How many different samples of size n = 3 can be selected from a population with the following sizes? (5pts) ​

(C) You're given a probability mass function,

The mean is

The variance is

The standard deviation is simply the square root of the variance:

(D) I'm not entirely sure what is being asked here, so I'm kinda guessing at the meaning. I think the question is saying there is a large set of 5000 test scores that are normally distributed with mean µ = 86 and standard deviation σ = 10.

Let X be the random variable representing these test scores. Then

(D.A)

where Z follows the standard normal distribution with mean 0 and variance 1.

To find the remaining probability, you can use the empirical rule (68/95/99.7) which says

• approximately 68% of a normal distribution lies within 1 standard deviation of the mean; in other words,

• approximately 95% of the distribution lies within 2 standard deviations;

The normal distribution is also symmetric about its mean. Taking these facts together, we find

So roughly 13.5% of all test scores will fall between 96 and 106, and 13.5% of 5000 is 675. (The actual probability is closer 0.135905, and the projected test score count is closer to 679.)

(D.B) Any 50% of the distribution is still 50% of the distribution, so half of all the test scores would fall in this range. There would be 2500 test scores in that group.

(E) No choices given here...

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step-by-step explanation:

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34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm