Mathematics, 18.05.2021 19:10, chanthonymjones

A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 4X2)/8. Determine the bias and the mean squared error for this estimator of the mean. Bias: Mean Square Error:

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answered: maddie53116

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

$\mu = \frac{3X_{1} + 4X_{2} }{8}$

Now

Bias for the estimator = E(μ bar) - μ

= $E( \frac{3X_{1} + 4X_{2} }{8}) - 4.5$

= $\frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5$

= $\frac{3(4.5) + 4(4.5)}{8} - 4.5$

= $\frac{13.5 + 18}{8} - 4.5$

= $\frac{31.5}{8} - 4.5$

= 3.9375 - 4.5

= - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

= Var(μ bar) + [Bias(μ bar, μ)]²

= $Var( \frac{3X_{1} + 4X_{2} }{8}) + 0.3136$

= $\frac{1}{64} Var( {3X_{1} + 4X_{2} }) + 0.3136$

= $\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136$

= $\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136$

= $\frac{1}{64} [7(57.76)}] } + 0.3136$

= $\frac{1}{64} [404.32] } + 0.3136$

= 6.3175 + 0.3136

= 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

answered: Guest

Bc equals mb that should be your answer

answered: Guest

the arc length is found by finding the circumference of the full circle (2xpixradius) and then dividing by 4 to find just one quarter of the circle (90 degrees).

Which best describes how to find the length of an arc in a circle?

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A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get tw...