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2 years ago

3.

Mathematics

1 answer:

Nana76 [90]2 years ago

6 0

Answer:

no I do not agree, Because it doesn't make since

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What is 10 1/2 divided by 2 1/4

victus00 [196]

Answer:

4.666666667 or 4.6_ (repeating decimal)

7 0

2 years ago

You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e

victus00 [196]

You intend to estimate a population mean μ from the following sample. 26.2 27.7 8.6 3.8 11.6 You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places.

Answer:

The Confidence interval = (8.98 , 22.18)

Step-by-step explanation:

From the given information:

mean = $\dfrac{ 26.2+ 27.7+ 8.6+ 3.8 +11.6 }{5}$

mean = 15.58

the standard deviation $\sigma$ = $\sqrt{\dfrac{\sum(x_i - \mu)^2 }{n}}$

the standard deviation = $\sqrt{\dfrac{(26.2 - 15.58)^2 +(27.7 - 15.58)^2 +(8.6 - 15.58)^2 + (3.8 - 15.58)^2 + (11.6 - 15.58)^2 }{5 } }$

standard deviation = 9.62297

Degrees of freedom df = n-1

Degrees of freedom df = 5 - 1

Degrees of freedom df = 4

For df at 4 and 80% confidence level, the critical value t from t table = 1.533

The Margin of Error M.O.E = $t \times \dfrac{\sigma}{\sqrt{n}}$

The Margin of Error M.O.E = $1.533 \times \dfrac{9.62297}{\sqrt{5}}$

The Margin of Error M.O.E = $1.533 \times 4.3035$

The Margin of Error M.O.E = 6.60

The Confidence interval = ( $\mu \pm M.O.E$ )

The Confidence interval = ( $\mu + M.O.E$ , $\mu - M.O.E$ )

The Confidence interval = ( 15.58 - 6.60 , 15.58 + 6.60)

The Confidence interval = (8.98 , 22.18)

7 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

About 16 % of the population of a large country is allergic to pollen. If two people are randomly selected, what is the probabil

Vinvika [58]

Answer:

(a) 0.0256

(b) 0.2944

Step-by-step explanation:

For solving this exercise we can apply the binomial distribution. The equation that give as the probability is:

$P(x,n,p)=(nCx)*p^{x} *(1-p)^{n-x}$

Where n is the number of identical events, p is the probability that the event has a success and x is the number of success in the n identical events.

Additionally, nCx is calculated as:

$nCx=\frac{n!}{x!(n-x)!}$

So, in this case we have 2 identical events because we are going to select two people randomly and the probability p of success is the probability that the person is allergic to pollen and this probability is 16%.

Then, for the first case, x is 2 because their are asking for the probability that both are allergic to pollen. Replacing the values of x, n and p we get:

$P(2,2,0.16)=(2C2)*0.16^{2} *(1-0.16)^{2-2}$

P(2,2,0.16)=0.0256

For the second case, their are asking for the probability that at least one is allergic to pollen, that means that we need to sum the probability that both are allergic to pollen with the probability that just one is allergic to pollen.

Using the same equation to calculate P(1,2,0.16) we get:

$P(1,2,0.16)=(2C1)*0.16^{1} *(1-0.16)^{2-1}$