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3 years ago

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A statistician is collecting data to help her estimate the number of pickpockets in a certain city. she needs 150 pieces of data

and is 34 percent done. how many pieces of data has she collected?

1 answer:

ololo11 [35]3 years ago

7 0

51 pieces of data will be collected

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Pic attached for further detail

marissa [1.9K]

I dont get what you asking

8 0

3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of <em>k</em> is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of <em>k</em> is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

<u>Check condition 1:</u>

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

<u>Check condition 2:</u>

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

7 0

3 years ago

Prove that the two circles shown below are similar.

Setler [38]

Answer:

The scale factor is $\frac{R_{A}}{R_{C}}=\frac{5}{2}$

<u>We can say that both circles are similar.</u>

Step-by-step explanation:

If we move the little circle to the center of the bigger circle, so the <u>translate vector will be (3,5).</u>

Now we realize that the bigger circle is just a dilation of the smaller circle, the<u> scale factor is:</u>

$R_{A}=5$

$R_{C}=2$

$\frac{R_{A}}{R_{C}}=\frac{5}{2}$

Therefore, <u>we can say that both circles are similiar.</u>

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I hope it helps you!

8 0

3 years ago

When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for

juin [17]

Answer: 0.332 < p < 0.490

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

$\hat{p}$ = sample proportion

z* = critical z-value.

As per given , we have

n= 258

Sample proportion of college students who own a car = $\hat{p}=\dfrac{106}{258}\approx0.411$

Critical z-value for 99% confidence interval is 2.576. (By z-table)

Therefore , the 99% confidence interval for the true proportion(p) of all college students who own a car will be : $0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)$

Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490

3 0

3 years ago

BAD is congruent to BCD by <br> the

scoundrel [369]

AB is congruent to BC

6 0

3 years ago