Probability

The odds are... remote. This is a factorial problem to calculate the answer you take how many options the first "dealt" card has of being. In this case the first card has a 1-in-52 chance of being the Ace of Spades (the "highest" card in a deck.) The next card has a 1-in-51 chance of being the Two of Spades, and so on. You calculate the odds of this happening my multiplying 52 by 51.

So, the odds are 52 times 51 times 50 times 49 times 48 times... yadda yadda yadda until you get to one. The answer?

The odd of taking a shuffled deck of cards, reshuffling it, and the dealing the cards back in the correct order are 1 in 806,581,751,709,439x10 to the 54th power. (septendecillion.)

Or 80 unvigintillion. (80 followed by 66 zeroes). For the sake of comparison, conservative estimates put the number of stars in our galaxy at 100 Billion (100 followed by 9 zeroes) and the number of galaxies in the universe at the same number, making the number of potential stars in the universe at 10 sextilion (a 10 followed by 21 zeroes.)

A sample.

The answer will depend on whether the critical region is one-tailed or two-tailed.

The answer depends on how many cards are drawn.

Primes are 2, 3, 5 and 7 which form 6 "prime pairs". There are 45 possible pairs in the box so the odds are 39 to 6 against ie 13/2, a probability of two-fifteenths or 13.3% or 0.13 recurring.

The median is the most appropriate center when the distribution is very skewed or if there are many outliers.

true

z = 0.5244, approx.

0.674 sd.

0.3

If two fair number cubes are rolled, the answers are:

Prob(7) = 6/36 = 1/6

Prob(11) = 2/36 = 1/18.

The median and mode.

For 3 you need 1+2, but for 5 you can have 1+4 or 2+3 so you have twice as many chances for 5.

Empirical

No, a standard deviation or variance does not have a negative sign. The reason for this is that the deviations from the mean are squared in the formula. Deviations are squared to get rid of signs.

In Absolute mean deviation, sum of the deviations is taken ignoring the signs, but there is no justification for doing so. (deviations are not squared here)

main db using mcb sub main db using mccb that is deference

The standard normal table tells us the area under a normal curve to the left of a number z.

The tables usually give only the positive value since one can use symmetry to find the corresponding negative values.

The middle 60 percent leaves 20 percent on either side. So we want the z scores that correspond to that 80 percentile which is .804. Therefore the values are are between z scores of -.804 and .804

* * * * *

I make it -0.8416 to 0.8416

2.16

Every roll of a fair die has a one in two chance of being even or odd.

This could easily happen with a fair coin after only

There is 2 outcomes for flipping the coin, and 6 outcomes for rolling the cube. The total outcomes for both are 2*6 = 12.

The number of possible outcomes would be 2^6 = 64, since each of the six questions have two possible outcomes.

Because the marble has a higher density than the water does

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