Probability

Yes it is. Discrete data is something that's set. Like say you were making a line graph about renting bikes. You can only rent whole bikes there is nothing in between. You shouldn't connect points on a line graph with discrete data but some cases can be argued.

Continuous data is usually a measurement that could change like time.

25/36

64

Going by the assumption that the probability of a child being a girl is 1/2, and that the events are independent, then the probabilities of different numbers of girls are as follows:

P(0 girls) = 1/2 ^ 4 = 1/16

P(1 girl) = 4 * (1/2) * (1/2 ^ 3) = 1/4

P(2 girls) = 6 * (1/2 ^ 2) * (1/2 ^ 2) = 3/8

P(3 girls) = 4 * (1/2 ^ 3) * (1/2) = 1/4

P(4 girls) = 1/2 ^ 4 = 1/16

Therefore the probability of at least 2 girls = 3/8 + 1/4 + 1/16 = 11/16

The probability that both coins are heads is the probability of one coin landing heads multiplied by the probability of the second coin landing heads: (.5) * (.5) = .25 or (1/2) * (1/2) = 1/4

Mean 0, standard deviation 1.

The mean deviation or absolute mean deviation is the sum of the differences between data values and the mean, divided by the count. In this case the MAD is 6.

It would be 3*5 = 15.

It means the two events cannot simultaneously occur; for example the two events, being dead and being alive are mutually exclusive, since they cannot occur at the same time.

gregor mendel

True.

If the odd favoring an event are 10 to 1, then the probability of the event occurring is 0.9.

The odds in favor of an event are 10:1.

Find the probability that the event will occur.

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P(E)+P(E') = 1

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P(E)/P(E') = 10/1

So P(E) = 10P(E')

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Substitute for P(E) and solve for P(E'):

10P(E')+P(E') = 1

11P(E') = 1

P(E') = 1/11

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Therefore P(E) = 10/11

2 out of 3 times 3 out of 5

98.73

That depends on the size of the marbles, as well as the size and shape of the jar.

According to the central limit theorem, as the sample size gets larger, the sampling distribution becomes closer to the Gaussian (Normal) regardless of the distribution of the original population. Equivalently, the sampling distribution of the means of a number of samples also becomes closer to the Gaussian distribution. This is the justification for using the Gaussian distribution for statistical procedures such as estimation and hypothesis testing.

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

It is an ordered pair of the form (A, n) where A is the outcome of the tossed coin (H or T) and n is the outcome of the rolled die (1, 2, 3, 4, 5, 6).

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.