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2 years ago

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

3 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:

$P(X = 8) = 0.0037$

$P(X < 5) = 0.805$

$P(X > 6) = 0.0206$

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

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

$P(X = 8) = 0.0037$

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

$P(X > 6) = 1 - p(X \le 6)$