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ioda
3 years ago
14

Pls, can you help me? thx.

%20%5Cfrac%7B%20%5Csqrt%7B2%7D%20%7D%7B2%7D" id="TexFormula1" title=" \cos(x + \frac{\pi}{3}) \geqslant \frac{ \sqrt{2} }{2}" alt=" \cos(x + \frac{\pi}{3}) \geqslant \frac{ \sqrt{2} }{2}" align="absmiddle" class="latex-formula">​
Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

For x between -\pi and \pi, we have

  • \cos x=\frac{\sqrt2}2=\frac1{\sqrt2} when x=\pm\frac\pi4;
  • \cos x=1 for x=0; and
  • \cos x=0 for x=\pm\frac\pi2

\cos x is continuous over its domain, so the intermediate value theorem tells us that

\cos x\ge\frac{\sqrt2}2

is true for -\frac\pi4\le x\le\frac\pi4.

For all x, we take into account that \cos x is 2\pi-periodic, so the above inequality can be expanded to

-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4

where n is any integer. Equivalently,

-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi

To get the corresponding solution set for

\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2

simply replace x with x+\frac\pi3:

-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi

\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}

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Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

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

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

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

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

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

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

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