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3 years ago

6

The local amusement park was interested in the average wait time at their most popular roller coaster at the peak park time (2 p

.m.). They selected 13 patrons and had them get in line between 2 and 3 p.m. Each was given a stop watch to record the time they spent in line. The times recorded were as follows (in minutes)
118, 124, 108, 116, 99, 120, 148, 118, 119, 121, 45, 130, 118

What is the first quartile?
A. 100.8
B. 119.8
C. 128.8
D. 112
E. 122.5

1 answer:

hichkok12 [17]3 years ago

4 0

Answer:

(D) 112

Step-by-step explanation:

First step: Arrange all the numbers from in ascending order that is from smallest to largest.

In Ascending order the set becomes:

45,99,108,116,118,118,118,119,120,121,124,130,148.

Second step: Find the median of the set of numbers.

Median is the middle number.

The median is 118. The median is also called the second quartile(Q2).

Third step: Now we consider all the numbers to the Left of the median(Q2). The numbers are listed below:

45,99,108,116,118,118.

Now we will find the median of these numbers.

The total numbers are 6 which is an even number.

To find the median of an even number, we add the two middle numbers and divide them by 2.

The two middle numbers are 108 and 116.

Therefore 108+116 = 224.

224/2 =112.

Therefore 112 becomes the first quartile (Q1).

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If XX is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases:

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Answer:

(a) The mean, variance and standard deviation of <em>X </em>are 1.60, 0.96 and 0.98 respectively.

(b) The mean, variance and standard deviation of <em>X </em>are 3.20, 0.64 and 0.80 respectively.

(c) The mean, variance and standard deviation of <em>X </em>are 1.50, 0.75 and 0.87 respectively.

(d) The mean, variance and standard deviation of <em>X </em>are 4.00, 0.80 and 0.89 respectively.

Step-by-step explanation:

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The mean, variance and standard deviation of <em>X</em> are:

$\mu=np\\\sigma^{2}=np(1-p)\\\sigma=\sqrt{np(1-p)}$

(a)

For <em>n</em> = 4 and <em>p</em> = 0.40 compute the mean, variance and standard deviation of <em>X </em>as follows:

$\mu=np=4\times0.40=1.60\\\sigma^{2}=np(1-p)=4\times0.40\times(1-0.40)=0.96\\\sigma=\sqrt{np(1-p)}=\sqrt{0.96}=0.98$

Thus, the mean, variance and standard deviation of <em>X </em>are 1.60, 0.96 and 0.98 respectively.

(b)

For <em>n</em> = 4 and <em>p</em> = 0.80 compute the mean, variance and standard deviation of <em>X </em>as follows:

$\mu=np=4\times0.80=3.20\\\sigma^{2}=np(1-p)=4\times0.80\times(1-0.80)=0.64\\\sigma=\sqrt{np(1-p)}=\sqrt{0.64}=0.80$

Thus, the mean, variance and standard deviation of <em>X </em>are 3.20, 0.64 and 0.80 respectively.

(c)

For <em>n</em> = 3 and <em>p</em> = 0.50 compute the mean, variance and standard deviation of <em>X </em>as follows:

$\mu=np=3\times0.50=1.50\\\sigma^{2}=np(1-p)=3\times0.50\times(1-0.50)=0.75\\\sigma=\sqrt{np(1-p)}=\sqrt{0.75}=0.87$

Thus, the mean, variance and standard deviation of <em>X </em>are 1.50, 0.75 and 0.87 respectively.

(d)

For <em>n</em> = 5 and <em>p</em> = 0.80 compute the mean, variance and standard deviation of <em>X </em>as follows:

$\mu=np=5\times0.80=4.00\\\sigma^{2}=np(1-p)=5\times0.80\times(1-0.80)=0.80\\\sigma=\sqrt{np(1-p)}=\sqrt{0.80}=0.89$

Thus, the mean, variance and standard deviation of <em>X </em>are 4.00, 0.80 and 0.89 respectively.

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3 years ago

The commute time for people in a city has an exponential distribution with an average of 0.5 hours. What is the probability that

mamaluj [8]

Answer:

0.314 = 31.4% probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this question:

$m = 0.5, \mu = \frac{1}{0.5} = 2$

What is the probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours?

$P(0.4 \leq X \leq 1) = P(X \leq 1) - P(X \leq 0.4)$

In which

$P(X \leq 1) = 1 - e^{-2} = 0.8647$

$P(X \leq 0.4) = 1 - e^{-2*0.4} = 0.5507$

So

$P(0.4 \leq X \leq 1) = P(X \leq 1) - P(X \leq 0.4) = 0.8647 - 0.5507 = 0.314$

0.314 = 31.4% probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours

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