Mathematics, 12.03.2021 22:20 Roquelle6301

A company manufactures a brand of lightbulb with a lifetime in months that is normally distributed with mean 3 and variance 1. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes. What is the smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772

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The smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772 is 16.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

n values from a normal distribution:

The mean is $\mu n$ and the standard deviation is $s = \sigma\sqrt{n}$

A company manufactures a brand of lightbulb with a lifetime in months that is normally distributed with mean 3 and variance 1.

This means that $\mu = 3, \sigma = \sqrt{1} = 1$

For n bulbs:

The distribution for the sum of n bulds has $\mu = 3n, \sigma = \sqrt{n}$

What is the smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772?

We want that: $S_{n} \geq 40 = 0.9772$ .

This means that when X = 40, Z has a pvalue of 1 - 0.9772 = 0.0228, that is, when X = 40, Z = -2 . So

$Z = \frac{X - \mu}{\sigma}$

$-2 = \frac{40 - 3n}{\sqrt{n}}$

$-2\sqrt{n} = 40 - 3n$

$3n - 2\sqrt{n} - 40 = 0$

Using $y = \sqrt{n}$

3y^2 - 2y - 40 = 0

Which is a quadratic equation with a = 3, b = -2, y = -40

$\Delta = b^{2} - 4ac = (-2)^2 - 4(3)(-40) = 484$

$y_{1} = \frac{-(-2) + \sqrt{484}}{2*3} = 4$

$y_{2} = \frac{-(-2) - \sqrt{484}}{2*3} = -...$

Since y and n have both to be positive:

$y = \sqrt{n}$

$\sqrt{n} = 4$

$(\sqrt{n})^2 = 4^2$

n = 16

The smallest number of bulbs to be purchased so that the succession of bulbs produces light for at least 40 months with probability at least 0.9772 is 16.

Answer from: Quest

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

\/ \/ \/ \/

48 + 24 - 408 + 24 - 3

\ / \ /

72 - 432 - 3 = -363

Answer from: Quest

answer: 1722

step-by-step explanation:

given: the mass of saturn= $5.685\times10^{26}\ kgs$

the mass of mercury= $3.302\times10^{23}\ kgs$

the number of times the mass of saturn is greater than the mass of mercury

$=\frac{\text{mass of saturn}}{\text{mass of mercury}}\\=\frac{5.685\times10^{26}}{3.302\times10^{23}}\\\\=1.72168\times10^{26-23}{by using law of exponent}]\\=1.72168\times10^3=1721.68\approx1722$

hence, the mass of saturn is 1722 times greater than the mass of mercury.

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