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2 years ago

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The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the act

ual temperature of the medium, and standard deviation σ. What would the value of σ have to be to ensure that 95% of all readings are within 0.5° of μ? (Round your answer to four decimal places.) σ =

1 answer:

stealth61 [152]2 years ago

3 0

Answer: 0.2551

Step-by-step explanation:

Given : The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the actual temperature of the medium, and standard deviation σ.

Significance level : $\alpha=1-0.95=0.05$

The critical z-value for 95% confidence : $z_{\alpha/2}=1.960$ (1)

Since , $z=\dfrac{x-\mu}{\sigma}$ (where x be any random variable that represents the temperature reading from a thermocouple.)

Then, from (1)

$\dfrac{x-\mu}{\sigma}=1.96\\\\ x-\mu=1.96\sigma$ (2)

Also, all readings are within 0.5° of μ,

i.e. $x-\mu$

i.e. $1.96\sigma$ [From (2)]

i.e. $\sigma$

i.e. $\sigma\approx0.2551$

The required standard deviation : $\sigma=0.2551$

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

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Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

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$\Rightarrow A= 92 \times 2$

⇒A= 184.

Therefore $k= \frac{ln2 }{18}$ , A=184

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Seems to be that the limit to compute is

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Answer:

The <em>z</em>-score for the group "25 to 34" is 0.37 and the <em>z</em>-score for the group "45 to 54" is 0.25.

Step-by-step explanation:

The data provided is as follows:

25 to 34 45 to 54

1329 2268

1906 1965

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1826 1591

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Compute the mean and standard deviation for the group "25 to 34" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [1329+1906+...+1454]=\frac{13631}{8}=1703.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1086710.875}=394.01$

Compute the <em>z</em>-score for the group "25 to 34" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1703.875}{394.01}=0.3734\approx 0.37$

Compute the mean and standard deviation for the group "45 to 54" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [2268+1965+...+2158]=\frac{14031}{8}=1753.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1028888.875}=383.39$

Compute the <em>z</em>-score for the group "45 to 54" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1753.875}{383.39}=0.25333\approx 0.25$

Thus, the <em>z</em>-score for the group "25 to 34" is 0.37 and the <em>z</em>-score for the group "45 to 54" is 0.25.

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A geometric series is written as $ar^n$ , where $a$ is the first term of the series and $r$ is the common ratio.

In other words, to compute the next term in the series you have to multiply the previous one by $r$ .

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

$6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots$ .

Let's use the other information, since the last term is $4374 > 6$ , we know that $r>1$ , otherwise the terms would be bigger and bigger.

The information about the sum tells us that

$\displaystyle \sum_{i=0}^n 6r^i = 6\sum_{i=0}^n r^i = 6558$

We have a formula to compute the sum of the powers of a certain variable, namely

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So, the equation becomes

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