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PtichkaEL [24]
3 years ago
9

3850 times a 1.3% interest rate for 6 years

Mathematics
1 answer:
netineya [11]3 years ago
7 0

"sam places 3850 in a bank account that pays 1.3% simple interest per year and how much interest will she earn in 6 years | Wyzant Ask An Expert" https://www.wyzant.com/resources/answers/646803/sam-places-3850-in-a-bank-account-that-pays-1-3-simple-interest-per-year-an3850

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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
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Answer:

(a) The value of P (X ≤ 2) is 0.8729.

(b) The value of P (X ≥ 5) is 0.0072.

(c) The value of P (1 ≤ X ≤ 4) is 0.7154.

(d) The probability that none of the 25 boards is defective is 0.2774.

(e) The expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defective boards.

The probability that a circuit board is defective is, <em>p</em> = 0.05.

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

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...

(a)

Compute the value of P (X ≤ 2) as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

P(X\leq =x)=\sum\limits^{2}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.2774+0.3650+0.2305\\=0.8729

Thus, the value of P (X ≤ 2) is 0.8729.

(b)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = 1 - P (X < 5)

              =1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072

Thus, the value of P (X ≥ 5) is 0.0072.

(c)

Compute the value of P (1 ≤ X ≤ 4) as follows:

P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

                   =\sum\limits^{4}_{x=1}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.3650+0.2305+0.0930+0.0269\\=0.7154

Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.

(d)

Compute the value of P (X = 0) as follows:

P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774

Thus, the probability that none of the 25 boards is defective is 0.2774.

(e)

Compute the expected value of <em>X</em> as follows:

E(X)=np=25\times 0.05=1.25

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09

Thus, the expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.

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