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Svetradugi [14.3K]
3 years ago
12

Problem 4. (3 points) Given the following matrix A = 1 1 0 1 . Which of the following statements must be TRUE? (I) The rank of t

he matrix is one. (II) The matrix A is not diagonalizable. (III) The nullity of A is one. (A) I only (B) II only (C) III only (D) II and III only (E) I, II, and III
Mathematics
1 answer:
shepuryov [24]3 years ago
8 0

Answer:

C) II only

Step-by-step explanation:

Given the 2×2 matrix A = [1 1, 0 1]

The only statement true there is (II) only i.e the matrix is not diagonalizable but invertible. A matrix is invertible if the product of the matrix and its inverse is equal to an identity matrix

The rank of the matrix is not 1 but 2 because rank of a matrix is the number of non zero rows of a matrix and the number of non zero rows of this matrix is 2 thereby making I incorrect.

Also note that the sum of nullity and rank is equal to the number of columns of the given matrix

Nullity + rank = number of columns

Nullity = number of columns - rank

Nullity = 2-2 = 0

The nullity is therefore 0 not 1 making option III also false

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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a
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(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let <em>S</em> = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = <em>p</em> = 0.30.

The number of sample selected is, <em>n</em> = 25.

The random variable S\sim Bin(25,0.30)

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion (\hat p) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: E(\hat p)=p=0.30

The standard deviation of the the sampling distribution of the sample proportion is:

SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: p=\frac{5}{25} =0.20

P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: p=\frac{7}{25} =0.28

P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: p=\frac{9}{25} =0.36

P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: p=\frac{11}{25} =0.44

P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq  \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558

The exact probability is more than the approximated probability.

6 0
3 years ago
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