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lara [203]
4 years ago
5

Randy spins the arrow on a spinner with 5 equal sections labeled A,B,C,D, and EVEN. Then ,he rolls a 6-sided number cube with si

des numbered 1 through 6. What is the problability that the arrow will stop on the letter A and the number cube will show. The number 4. I have no idea how to do this
Mathematics
2 answers:
aivan3 [116]4 years ago
8 0
You have 1 out of 5 chances to get a A in the spinner 
<span>You have a 1 out of 6 chances to get a 4 </span>
<span>1/5 times (x) 1/6 = 1/30</span>
Rufina [12.5K]4 years ago
5 0
Hello there.

Question: <span>Randy spins the arrow on a spinner with 5 equal sections labeled A,B,C,D, and EVEN. Then ,he rolls a 6-sided number cube with sides numbered 1 through 6. What is the probability that the arrow will stop on the letter A and the number cube will show. The number 4.

Answer: There is a 1/5 chance of getting A.
There is a 1/6 chance of getting a 4.

</span>Hope This Helps You!
Good Luck Studying ^-^
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Answer:

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

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                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

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