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1. Find the mean; the variance; and the probability that random variable falls between -1.6 and 2.8, if the mgf of the variable is M(t) (est 3t) for #0. and M(0) = ...

Question

1. Find the mean; the variance; and the probability that random variable falls between -1.6 and 2.8, if the mgf of the variable is M(t) (est 3t) for #0. and M(0) = 1. b) M(t) = e-3t+5t?_ ~0 < t < 0.

1. Find the mean; the variance; and the probability that random variable falls between -1.6 and 2.8, if the mgf of the variable is M(t) (est 3t) for #0. and M(0) = 1. b) M(t) = e-3t+5t?_ ~0 < t < 0.



Answers

Find the mean and variance of the normal distribution of statistics using parts (a) and (b) with $m(t)=e^{\mu t+\sigma^{2} t^{2} / 2}.$ (a) Mean $=m^{\prime}(0)$ (b) Variance $=m^{\prime \prime}(0)-\left(m^{\prime}(0)\right)^{2}$

So in this question, we're told that a population has a mean of 16 and a standard deviation 1.7. And first we asked for the mean and standard deviation of the sample means, so that's mean of the sample mean is just the same as the population mean. The standard deviation of the sample mean is standard deviation of the population mean over square root of the sample size, which is 1.7 over Route 80, which is .1901. But we were asked for the probability that a mean of a sample of size 80 Will be more than 16.4. So we have the probability that are standard moment when the variable is more than 13.4 -16 over .1901. So That value here 16.4 -16 Over our standard deviation. So that value is 2.1 zero Or 2.11. So that value 2.10 which is basically one minus the probability Z is less than 2.10 So z being less than 2.10 is .98-1, So 1 -1821 is .017, no

Probleble 18. We want to identify this probability density function over the center, then find the mean the variance and the standard division. Without integration. We can rewrite F of X to be in the form one Divided by six. Multiply. School route of two boy buoyed by E. It's about -X -30. All squared divided by Two multiplied by six square. Because two multiplied by six square is to multiply by 36 gives us 17. Then F of X is in the 41 divided by segment square root. Two boy employed by E to the bar of minus x minus mu old square divided by two multiplied by sigma squared. And this and terrible means X is between minus infinity and infinity. This is a form of the normal probability density function. This means if it's normal for the normal distribution um you is given here is mu and sigma is given here. His sigma then new equals 13 and sigma equals six and the variance equals sigma squared equals six squared equals 36. Then this is a final answer of our

Okay, So for this question, we're gonna notice that our density function here you to the negative t over three, divided by three is exactly the form of an exponential distribution with Lambda equals 1/3. So if we want to because this is an exponential distribution. We have these formulas down here that we can make use off. So if we want to find the expectation of T, which is the mean we know, it's just one over Lambda. So that will be three If we want to find the standard deviation. We know that's just the square root of the variants which will be one over lambda again. So three again. And if we want to find the probability that, uh, rt is within one standard deviation of its mean, uh from above, that's going to be the probability that T is between three and three plus three. So between three and six and that follows exactly this formula down here. So we know that that's going to be 0.23 25 So these are our final answers


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