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The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p

Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

3 0

3 years ago

10x –3 = 4x + 8 What Is X?

navik [9.2K]

Answer:

x=1.833

Step-by-step explanation:

7 0

3 years ago

The bottom of a ladder must be placed 4 feet from the base of a pole barn. The ladder is 14 feet long. How far above the ground

marusya05 [52]

Check the picture below.

3 0

3 years ago

While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta

Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean $\mu$ and standard deviation 0.68 in dollars.

We average $n=45$ samples to get $\bar{x}$ .

The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write

$\sigma = 0.68 / \sqrt{45} = 0.101$

Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains $\mu$ .

Our interval takes the form of $( \bar{x} - z \sigma, \bar{x} + z \sigma )$ as $\bar{x}$ is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".

Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.

Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.

With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.

We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is

$( \bar{x} - 1.65 (.101), \bar{x} + 1.65 (.101) )$