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melomori [17]
2 years ago
14

9

Mathematics
1 answer:
Gwar [14]2 years ago
6 0

Answer:

It is A and I am positive its A

Step-by-step explanation:

You might be interested in
What is the minimum number of clients the travel agent should survey? Note that z=1.96 for a 95% confidence interval.
slega [8]

Answer:

121

Step-by-step explanation:

Given data as per the question

Standard deviation = \sigma = 840

Margin of error = E = 150

Confidence level = c = 95%

For 95% confidence, z = 1.96

based on the above information, the minimum number of clients surveyed by the travel agent is

n = (\frac{z\times \sigma}{E})^2

=  (\frac{1.96\times 840}{150})^2

= 120.47

= 121

hence, the 121 number of clients to be surveyed

Therefore we applied the above formula to determine the minimum number of clients

4 0
3 years ago
There are 30 students on the school's student council. A special homecoming dance committee is to be formed by randomly selectin
Oksana_A [137]
There can be a possible of 5 committees 
7 0
2 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
2 years ago
According to a study done by a university​ student, the probability a randomly selected individual will not cover his or her mou
VikaD [51]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 P(X = 8) =  0.0037

b

 P(X <  5) =  0.805

c

 P(X > 6) =  0.0206

I would be surprised because the value is very small , less the 0.05

Step-by-step explanation:

From the question we are told that

The probability a randomly selected individual will not cover his or her mouth when sneezing is p = 0.267

Generally data collected from this study follows  binomial  distribution because the number of trials is  finite , there are only two outcomes, (covering  , and  not covering mouth when sneezing ) , the trial are independent

Hence for a randomly selected variable  X we have that  

   X \ \ \~ \ \ { B ( p , n )}

The probability distribution function for binomial  distribution is  

    P(X = x ) =  ^nC_x *  p^x *  (1 -p) ^{n-x}

Considering question a

Generally the  the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing is mathematically represented as

     P(X = 8) =  ^{12} C_8 *  (0.267)^8 *  (1- 0.267)^{12-8}

Here C denotes  combination

So

     P(X = 8) =  495  *  0.000025828 * 0.28867947

    P(X = 8) =  0.0037

Considering question b

Generally the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when​ sneezing is mathematically represented as

     P(X <  5 ) =[P(X = 0 ) + \cdots + P(X = 4)]

=>   P(X <  5 ) =[ ^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0} + \cdots +  ^{12} C_4 *  (0.267)^4 *  (1- 0.267)^{12-4} ]

=> P(X <  5 )  =  0.02406 +  0.10516 + 0.21067 + 0.25580 + 0.20964

=>  P(X <  5) =  0.805

Considering question c

Generally the probability that fewer than half(6) covered their mouth when​ sneezing(i.e the probability the greater than half do not cover their mouth when sneezing) is mathematically represented as

      P(X > 6) =  1 - p(X \le  6)

=>    P(X > 6) = 1 - [P(X = 0) + \cdots + P(X =6)]

=>    P(X > 6)=1 - [^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0}+ \cdots + ^{12} C_4 *  (0.267)^6 *  (1- 0.267)^{12-6} ]

=>    P(X > 6)= 1 - [0.02406 + \cdots + 0.0519 ]  

=>    P(X > 6) =  0.0206

I would be surprised because the value is very small , less the 0.05

8 0
3 years ago
What is the sum of the areas of circle C and circle D?
Yanka [14]

Answer:

98pi

Step-by-step explanation:

Area of circle: pi(r)^2

A1=pi(7)^2

A2=pi(7)^2

A2+A1=98pi

4 0
2 years ago
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