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3 years ago

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In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Excel to fin

d the 90% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Round your answers totwo decimal places and use increasing order. Temperature 151 158 163 155 155 144 140 148 145 146 158 143 152 167 163 147 170 168 176 164

1 answer:

forsale [732]3 years ago

7 0

Answer:

The confidence interval is (152.03, 159.77)

Step-by-step explanation:

The formula for calculating the confidence interval for a population mean is given by:

$\bar{x}\pm \frac{z_{\alpha/2}*s}{\sqrt{n}}$

The sample size is actually $n = 20$ .

The sample average is $\bar{x}=155.90$ .

Using excel to calculate the standard deviation we get $s = 10.52$

The confidence level is $1-\alpha=0.90$ therefore $\alpha=0.10$

We obtain the critical value $z_{\alpha/2}=1.65$

In Excel I calculated the margin error before calculating the confidence interval. The margin error is given by:

$\frac{z_{\alpha/2}*s}{\sqrt{n}}=\frac{1.65*10.52}{\sqrt{20}}=3.87$

Now we can calculate the confidence interval.

$\bar{x}\pm \frac{z_{\alpha/2}*s}{\sqrt{n}}=155.90\pm3.87=(152.03, 159.77)$

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Let N be the smallest positive integer whose sum of its digits is 2021. What is the sum of the digits of N + 2021?

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Answer:

$10$ .

Step-by-step explanation:

See below for a proof of why all but the first digit of this $N$ must be " $9$ ".

Taking that lemma as a fact, assume that there are $x$ digits in $N$ after the first digit, $\text{A}$ :

$N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}$ , where $x$ is a positive integer.

Sum of these digits:

$\text{A} + 9\, x= 2021$ .

Since $\text{A}$ is a digit, it must be an integer between $0$ and $9$ . The only possible value that would ensure $\text{A} + 9\, x= 2021$ is $\text{A} = 5$ and $x = 224$ .

Therefore:

$N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}$ .

$N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}$ .

$N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}$ .

Hence, the sum of the digits of $(N + 2021)$ would be $6 + 2 + 2 = 10$ .

Lemma: all digits of this $N$ other than the first digit must be " $9$ ".

Proof:

The question assumes that $N\!$ is the smallest positive integer whose sum of digits is $2021$ . Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of $N$ is not " $9$ ".

For example: $N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}$ , where $\text{A}$ , $\text{P}$ , $\text{B}$ , and $\text{C}$ are digits. (It is easy to show that $N$ contains at least $5$ digits.) Assume that $\text{B} \!$ is one of the non-leading non-" $9$ " digits.

Either of the following must be true:

$\text{P}$ , the digit in front of $\text{B}$ is a " $0$ ", or
$\text{P}$ , the digit in front of $\text{B}$ is not a " $0$ ".

If $\text{P}$ , the digit in front of $\text{B}$ , is a " $0$ ", then let $N^{\prime}$ be $N$ with that " $0\!$ " digit deleted: $N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}$ .

The digits of $N^{\prime}$ would still add up to $2021$ :

$\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}$ .

However, with one fewer digit, $N^{\prime} < N$ . This observation would contradict the assumption that $N\!$ is the smallest positive integer whose digits add up to $2021\!$ .

On the other hand, if $\text{P}$ , the digit in front of $\text{B}$ , is not " $0$ ", then $(\text{P} - 1)$ would still be a digit.

Since $\text{B}$ is not the digit $9$ , $(\text{B} + 1)$ would also be a digit.

let $N^{\prime}$ be $N$ with digit $\text{P}$ replaced with $(\text{P} - 1)$ , and $\text{B}$ replaced with $(\text{B} + 1)$ : $N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}$ .

The digits of $N^{\prime}$ would still add up to $2021$ :

$\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}$ .

However, with a smaller digit in place of $\text{P}$ , $N^{\prime} < N$ . This observation would also contradict the assumption that $N\!$ is the smallest positive integer whose digits add up to $2021\!$ .

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this $N$ other than the first digit must be " $9$ ".

Therefore, $N$ would be in the form: $N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}$ , where $\text{A}$ , the leading digit, could also be $9$ .

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