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2 years ago

What is the simple interest rate on a $2100 investment paying $1,559.46 interest in 15.8 years?

Mathematics

1 answer:

cluponka [151]2 years ago

8 0

The simple interest rate is 4.7%

<h3>How to determine the simple interest rate?</h3>

The given parameters are:

Principal = $2100

Interest = $1,559.46

Time = 15.8 years

The simple interest is calculated as:

I = PRT

Make R the subject

R = I/(PT)

So, we have:

R = 1559.46/(2100 * 15.8)

Evaluate

R = 0.047

Express as percentage

R = 4.7%

Hence, the simple interest rate is 4.7%

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The 95% confidence interval for the true proportion would be given by (0.570;0.630) .

And if we convert this into % we got (57.0%, 63.0%)

Step-by-step explanation:

The information given we have the following info given:

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The confidence level is 95%, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

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The confidence interval for the mean is given by the following formula:

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Replacing the info given we got:

$0.60 - 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.570$

$0.60 + 1.96\sqrt{\frac{0.60(1-0.60)}{1019}}=0.630$

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3 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

andriy [413]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

(c) Standard deviation of defective light bulbs = 3.67

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

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<u>Standard deviation of X</u>, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

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Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

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P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

Standard deviation of defective light bulbs is given by = S.D. = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.67

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