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Let X be a binomial random variable with p=0.7, n=10. Find P(X =4); P(X > 12)....

Question

Let X be a binomial random variable with p=0.7, n=10. Find P(X =4); P(X > 12).

Let X be a binomial random variable with p=0.7, n=10. Find P(X = 4); P(X > 12).



Answers

A binomial probability experiment is conducted with the given parameters. Compute the probability of $x$ successes in the $n$ independent trials of the experiment. $$n=12, p=0.35, x \leq 4$$

And this question we are given to probabilities And we're asked to find the value that use the probability shown where access a normally distributed random variable mean 54 standard deviation 12. So we're going to use the same areas here in our conversion. So we're basically taking quality easy less than the stars equals two point goal in nine. And if we look at our table we get a value of Z star if he calls to minus 1.34 so X star, it's basically mu plus Z sigma, which gives us 37.92 458 would be for a given product X more than Next door's of poverty X. Less than X star is basically one minus this, which is 0.35 The value that's closest in RZ table This .3483 and RZ star values -199 And converting that to our extra value. Using this formula, we get 39.3 true.

In this problem, we need to determine the probability of X successes in the an independent trials of a binomial probability experiment. Now we need to return on the probability of the number of successes being less than or equal to four. So this will be the probability of the number of successes being either zero or 1 Or two or 3 or four. Now all of these outcomes are disjointed. So using the addition law, probability this will be P0 must be one The speech to let's be three, let's be four. Now, using the formula for the probability of X successes and a binomial probability experiment, the value of P zero with the N C X at 12 0 times P, which is 0.35 to the power X. Which is zero times one minus P. So one minus 0.350 point 65 Keep our end minus X. So that Australian minus zero with just 12. Similarly, P one is 12 C 10.35 to the power 10.65 to the power 11, P two will be 12 C 20.35 to the power to time 0.65 to the power 10, P three is 12 C 30.35 to the power 30.65 to the Power nine. And P four will be 12 C 40.35 to the power four times 0.65 to the power eight. Now the value of this will be approximately 0.00 5 7 plus 0.0368 plus 0.1088 Plus 0.19 by four Plus 0.2367. And the value of this is equal to 0.5834. Hence the required probability is approximately equal to 0.5834.

So we're trying to find the sum of all X squared times, probability of backs, and we're only given the sample size and the probability we can't solve this directly. But we can manipulate some equations of this chapter to find it, cause if you notice the thing we're looking for, this is part of the equation for variants, and we can find the mean directly, and we confined the variance for this equation. In the middle, the standard deviation is equal to a square root of variance. That means that standard deviation squared is equal toe. Very answer. So first, let's find the mean So we just multiply and and p together so 20 times point or that gives us eight the standard deviation equal to the square root of end times P, which is eight times one minus P. P's 10.4. So one minus point floor is 10.0.6, and this gives us 2.191 so standard deviation squared is equal to very answer, So 2.191 squared is equal to 4.8, so that means our variants is equal to 4.8. So I'll go over here now 4.8 equal to the thing we're looking for. Minus R means squared. So eight squared, which is 64. Add this to the over side We get 68.8 is equal to the sum of all X squared probability of X and we are done.

So we're giving. The probability is equal to one over end and X could be any number from one to end, and we're going to try and find the mean. So the equation for mean is the sum of all X times the probability of X Well, we have the probability of X that's one over end. So our probability of Axl's put it here one over and some of all X. So that is one plus two plus free all the way up to end. And the hint tells us what that is. Actually end times and close 1/2. Now, if you're wondering about the probability of X, it doesn't matter. Whether we include it in the summation or not, actually does not make a difference of So it's fine that we just plugged in one over and directly makes no difference. But from here, these two ends cancel out and we're just left with and plus 1/2, which is what we were trying to show. So really, you just needed this equation, and you just needed to notice that this some of X literally just means you add up every X value. But that is how you show that the mean is N plus 1/2


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