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2 years ago

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A card player selects a playing card from a well-shuffled deck of standard playing cards and sees if the card is a red card or n

ot. He keeps the card, and then makes another selection and sees if the card is a red card or not. He continues this process 10 times and counts the total number of red cards that he selects. Is this a binomial random variable? A. no since the total number of red cards in the deck is unknown at the start of the experiment B. no, since probabilities of drawing red cards do not remain the same from trial to trial C. yes since red cards are successes and he is counting the number of red cards drawn D. yes since a red card is a success and every other card is a failure

1 answer:

dmitriy555 [2]2 years ago

8 0

Answer:

A

Step-by-step explanation:

there could be no red cards inside

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Answer:

$(8967/n + 8.4)

Step-by-step explanation:

0.4nx + 0.6n(x+21) = 8967

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n = 300

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1. The following are the number of hours that 10 police officers have spent being trained in how to handle encounters with peopl

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Answer:

$Range = 16$

$Inter\ Quartile\ Range = 6.75$

$Variance = 20.44$

$Standard\ Deviation = 4.52$

Step-by-step explanation:

Given

4, 17, 12, 9, 6, 10, 1, 5, 9, 3

Calculating the range;

$Range = Highest - Lowest$

From the given data;

Highest = 17 and Lowest = 1

Hence;

$Range = 17 - 1$

$Range = 16$

Calculating the Inter-quartile Range

Inter quartile range (IQR) is calculates as thus

$IQR = Q_3 - Q_1$

Where

Q3 = Upper Quartile and Q1 = Lower Quartile

<em />

<em>Start by arranging the data in ascending order</em>

1, 3, 4, 5, 6, 9, 9, 10, 12, 17

N = Number of data; N = 10

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Calculating Q3

$Q_3 = \frac{3}{4}(N+1) th\ item$

<em>Substitute 10 for N</em>

$Q_3 = \frac{3}{4}(10+1) th\ item$

$Q_3 = \frac{3}{4}(11) th\ item$

$Q_3 = \frac{33}{4} th\ item$

$Q_3 = 8.25 th\ item$

Express 8.25 as 8 + 0.25

$Q_3 = (8 + 0.25) th\ item$

$Q_3 = 8th\ item + 0.25 th\ item$

Express 0.25 as fraction

$Q_3 = 8th\ item +\frac{1}{4} th\ item$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

From the arranged data;

$8th\ item = 10$ and $9th\ item = 12$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

$Q_3 = 10 +\frac{1}{4} (12 - 10)$

$Q_3 = 10 +\frac{1}{4} (2)$

$Q_3 = 10 +0.5$

$Q_3 = 10.5$

Calculating Q1

$Q_1 = \frac{1}{4}(N+1) th\ item$

<em>Substitute 10 for N</em>

$Q_1 = \frac{1}{4}(10+1) th\ item$

$Q_1 = \frac{1}{4}(11) th\ item$

$Q_1 = \frac{11}{4} th\ item$

$Q_1 = 2.75 th\ item$

Express 2.75 as 2 + 0.75

$Q_1 = (2 + 0.75) th\ item$

$Q_1 = 2nd\ item + 0.75 th\ item$

Express 0.75 as fraction

$Q_1 = 2nd\ item +\frac{3}{4} th\ item$

$Q_1 = 2nd\ item +\frac{3}{4} (3rd\ item - 2nd\ item)$

From the arranged data;

$2nd\ item = 3$ and $3rd\ item = 4$

$Q_1 = 3 +\frac{3}{4} (4 - 3)$

$Q_1 = 3 +\frac{3}{4} (1)$

$Q_1 = 3 +0.75$

$Q_1 = 3 .75$

---------------------------------------------------------------------------------

Recall that

$IQR = Q_3 - Q_1$

$IQR = 10.5 - 3.75$

$IQR = 6.75$

Calculating Variance

Start by calculating the mean

$Mean = \frac{1+3+4+5+6+9+9+10+12+17}{10}$

$Mean = \frac{76}{10}$

$Mean = 7.6$

Subtract the mean from each data, then square the result

$(1 - 7.6)^2 = (-6.6)^2 = 43.56$

$(3 - 7.6)^2 = (-4.6)^2 = 21.16$

$(4 - 7.6)^2 = (-3.6)^2 = 12.96$

$(5 - 7.6)^2 = (-2.6)^2 = 6.76$

$(6 - 7.6)^2 = (-1.6)^2 = 2.56$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(10 - 7.6)^2 = (2.4)^2 = 5.76$

$(12 - 7.6)^2 = (4.4)^2 = 19.36$

$(17 - 7.6)^2 = (9.4)^2 = 88.36$

Sum the result

$43.56 + 21.16 + 12.96 + 6.76 + 2.56 + 1.96 + 1.96 + 5.76 + 19.36 + 88.36 = 204.4$

Divide by number of observation;

$Variance = \frac{204.4}{10}$

$Variance = 20.44$

Calculating Standard Deviation (SD)

$SD = \sqrt{Variance}$

$SD = \sqrt{20.44}$

$SD = 4.52$ <em>(Approximated)</em>

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