Construct 95% fiducial limits of populations. Differentiate null and alternative hypothesis. Random variable X has the following probability distribution. Find the value of a. Find... Construct 95% fiducial limits of populations. Differentiate null and alternative hypothesis. Random variable X has the following probability distribution. Find the value of a. Find P(X<5), P(5<X<7). Show that A^2-2A-5I=0 where O is the 2x2 identity matrix.
Understand the Problem
The question involves a probability distribution where the probabilities are expressed in terms of a variable 'a'. It asks the user to find the value of 'a', as well as to calculate specific probabilities relating to the random variable X. Additionally, the question includes a matrix operation that requires verification of a matrix equation.
Answer
The value of \( a \) is \( \frac{1}{66} \), \( P(X<5) = \frac{25}{66} \), \( P(5<X<7) = \frac{1}{6} \).
Answer for screen readers
The value of ( a ) is ( \frac{1}{66} ), ( P(X<5) = \frac{25}{66} ), ( P(5<X<7) = \frac{1}{6} ).
Steps to Solve
- Find the value of ( a )
The probabilities must sum to 1. We set up the equation: $$ P(X) = a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1 $$
Combining the coefficients gives: $$ 66a = 1 $$
Solving for ( a ): $$ a = \frac{1}{66} $$
- Calculate ( P(X<5) )
To find ( P(X < 5) ), we add the probabilities for ( X = 0, 1, 2, 3, 4 ): $$ P(X<5) = P(0) + P(1) + P(2) + P(3) + P(4) $$ Substituting the values: $$ = a + 3a + 5a + 7a + 9a = (1 + 3 + 5 + 7 + 9)a = 25a $$ Thus, $$ P(X<5) = 25 \cdot \frac{1}{66} = \frac{25}{66} $$
- Calculate ( P(5<X<7) )
To find ( P(5 < X < 7) ), we consider ( X = 6 ): $$ P(5<X<7) = P(6) = 11a $$ Substituting ( a ): $$ = 11 \cdot \frac{1}{66} = \frac{11}{66} = \frac{1}{6} $$
- Verify the matrix equation ( A^2 - 2A - 5I = 0 )
Let $$ A = \begin{bmatrix} 1 & 2 \ 3 & 1 \end{bmatrix} $$ Calculating ( A^2 ): $$ A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot1 + 2\cdot3 & 1\cdot2 + 2\cdot1 \ 3\cdot1 + 1\cdot3 & 3\cdot2 + 1\cdot1 \end{bmatrix} = \begin{bmatrix} 7 & 4 \ 6 & 7 \end{bmatrix} $$
Now calculate ( -2A ): $$ -2A = -2\begin{bmatrix} 1 & 2 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & -4 \ -6 & -2 \end{bmatrix} $$
And ( -5I ): $$ -5I = -5 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 0 \ 0 & -5 \end{bmatrix} $$
Now combine: $$ A^2 - 2A - 5I = \begin{bmatrix} 7 & 4 \ 6 & 7 \end{bmatrix} + \begin{bmatrix} -2 & -4 \ -6 & -2 \end{bmatrix} + \begin{bmatrix} -5 & 0 \ 0 & -5 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} $$
This shows: $$ A^2 - 2A - 5I = 0 $$
The value of ( a ) is ( \frac{1}{66} ), ( P(X<5) = \frac{25}{66} ), ( P(5<X<7) = \frac{1}{6} ).
More Information
The calculations confirm that the probability distribution is valid as it sums to 1. The matrix verification demonstrates the properties of matrix operations and provides insights into linear algebra.
Tips
- Forgetting to sum all probabilities to equal 1 can lead to finding incorrect values of ( a ).
- Errors in matrix multiplication can occur if the dimensions are not checked correctly.
- Not carefully distributing coefficients in probability calculations can result in inaccurate probabilities.
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