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2 years ago

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Two types of flares are tested and their burning times are recorded. The summary statistics are given below: n = 35 n = 40 = 19.

4 min = 15.1 min s = 1.4 min s = 0.8 min Construct a 95% confidence interval for the differences between the mean burning time of the brand X flare and the mean burning time of the brand Y flare.
a. 3.2 min < ?X - ?Y < 5.4 min
b. 3.6 min < ?X - ?Y < 5.0 min
c. 3.8 min < ?X - ?Y < 4.8 min
d. 3.5 min < ?X - ?Y < 5.1 min

1 answer:

marusya05 [52]2 years ago

5 0

Answer:

C. 3.8 min < ?X - ?Y < 4.8 min

Step-by-step explanation:

Let both flares be X and Y

For X

n1 = 35

Bar x1 =19.4

D1 = 1.4

For Y

n2 = 40

Bar X2= 15.1

S2 = 0.8

√(x1²/n1) + (x2²/n2)

=√ (1.4²/35)+(0.8²+40)

= 0.2683

Critical value, t = 2.032

We now have to calculate margin of error

0.2683x2.032

= 0.545 this is approximately equal to 0.5

Bar x1 - bar X2

= 19.4 - 15.1

= 4.3

At 95% confidence level

4.3+-0.5

4.3+0.5= 4.8

4.3-0.5 = 3.8

Therefore the answer is 3.8min<X-Y<4.8min

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Step-by-step explanation:

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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t

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Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A bank gets an average of 12 customers per hour.

This means that $\mu = 12$

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

$P(X < 10) + P(X \geq 10) = 1$

We want $P(X \geq 10)$

Then

$P(X \geq 10) = 1 - P(X < 10)$

In which

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

$P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437$

$P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655$

$P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874$

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0 + 0.0001 + 0.0004 + 0.0018 + 0.0053 + 0.0127 + 0.0255 + 0.0437 + 0.0655 + 0.0874 = 0.2424$

Then

$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

75.76% probability that there will be 10 or more customers at this bank in one hour.

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Answer:

C. The probability that a randomly selected person who chose Mexico as the preferred destination is a child is $\frac{14}{113}$

Step-by-step explanation:

You are given a two-way table:

$\begin{array}{ccccc}&\text{Hawaii}&\text{Mexico}&\text{Florida}&\text{Total}\\\text{Child (less than13 years old)}&33&14&62&109\\\text{Teenager (13 - 17 years old)}&50&42&25&117\\\text{Adult (18 years old and up)}&64&57&8&129\\\text{Total}&147&113&95&355\end{array}$

Check all options:

A. The probability that a randomly selected adult chose Hawaii as the preferred destination.

Number of people who choose Hawaii = 147

Number of teenagers who choose Hawaii = 50

Probability is

$\dfrac{50}{147}$

False

B. The probability that a randomly selected child chose Florida as the preferred destination.

Number of childs = 109

Number of childs who prefer Florida = 62

The probability is

$\dfrac{62}{109}$

False

C. The probability that a randomly selected person who chose Mexico as the preferred destination is a child.

Number of people who choose Mexico = 113

Number of people who choose Mexico and is a child = 14

The probability is

$\dfrac{14}{113}$

True

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