If it took 30 minutes to give out 75 diplomas

Students at a virtual school are allowed to sign up for one math class each year. The numbers of students signing up for various

Margaret [11]

A well formatted table of the distribution is attached below :

Answer:

0.124

0.733

0.408

Step-by-step explanation:

Using the table Given :

1.) P(AP Statistics) = 90 / 725 = 0.124

2.) P(12th grade ; Precalculus or AP Statistics) = (100 + 65) / 225 = 165 /225 = 0.733

3.) P(Algebra 11 | 11th grade) = P(Algebara11 n 11th grade) / P(11th grade) = 100 / 245 = 0.408

7 0

2 years ago

Please help me :((((((((

romanna [79]

Answer:

Fraction Form: 1/10

Step-by-step explanation:

Decimal Form: 0.1

7 0

3 years ago

I need help please (brainllest)

topjm [15]

I think that the answer is B and D.

Both of them make sense because most of the time the variable X represents the key part to finding the solution. Also A lot of times there are more than one equations.

-Ɽ3₮Ɽ0 Ⱬ3Ɽ0

4 0

3 years ago

A diagnostic test for a certain disease is applied to n individuals known to not have the disease. Let X = the number among the

Anon25 [30]

Answer:

Part a

The maximum likelihood estimator of p, is the sample proportion, $\bar p=\frac{x}{n}$

And the proportion of the estimate at n=25 and x = 3 is 0.12.

Part b

The maximum likelihood estimate from part (a) is an unbiased estimator.

Part c

The maximum likelihood estimate of the probability is 0.5277.

Step-by-step explanation:

a)

The likelihood function is, $L(p;x) =\frac{n!}{x!(n-x)!}p*(1- p)^{n-x}$

Applying log on both sides:

$\log L(p;x) =\log\frac{n!}{x!(n-x)!}+x\log p+(n-x)log(1-p)$

Find the differentiation of log L with respect to P and equate to zero to find the MLE for .

$\frac{d}{dp}\log L(p,x)=\frac{x}{p}-\frac{n-x}{1-p}\\\\=\frac{x}{p}-\frac{(n-x)}{1-p}\\\\=x(1-p)-(n-x)p=0\\=x-xp-np+xp=0\\=x-np=0\\=x=np\\=\bar p=\frac{x}{n}$

Calculate the estimate of the function if n=25 and x = 3.

$\bar p=\frac{x}{n}\\\\=\frac{3}{25}\\\\= 0.12
$

Hint for next step

The MLE for the population proportion is the sample proportion and there is a 0.12 proportion of estimate at n = 25 and x = 3

b)

Consider

$E(\bar p)\\\\E(\bar p)=E(\frac{X}{n})\\\\=\frac{1}{n}E(X)\\\\=\frac{1}{n}(np)(since\, E(X)
=np)\\\\=p$

Since $E(\bar p)=p
$ , the maximum likelihood estimate from part (a) is an unbiased estimator.

Hint for next step

Based on the part (a) calculation, since it is proved that , the maximum likelihood estimate is unbiased estimator for population proportion

(c)

Calculate the maximum likelihood estimate of the probability $(1-p)^5$ .

By the Invariance property of MLE, to find the MLE of $(1-P)^5$ , the MLE of p can be used.

That is, the MLE of $(1-p)^5$ is $(1-\bar p)^5$

Therefore,

MLE of the probability,

$(1- p)^5 = (1-\bar P)^5\\\\=(1-\frac{3}{25})^5\\\\=(1-0.12)^5\\=(0.88)^5\\=0.5277$

7 0

3 years ago

550 people use a public swimming pool the prices are 1.75$ for children and 2.25$ for adults the receipts for admission totaled

natta225 [31]

Answer:

There were 169 children at the pool.

Step-by-step explanation:

Given:

Total number of people = 550

Price per child = $1.75

Price per adult = $2.25

Total cost of the admission receipts = $1153.00

Let the number of children be 'x' and number of adults be 'y'.

As per question:

Total number of people is equal to total children and total adults. So,

$x+y=550-----------(1)$

Total cost of admission is equal to the sum of cost of total children and total adults. So,

$1.75x+2.25y=1153-----------(2)$

Multiplying equation (1) by 1.75 and subtracting the result equation from equation (2). This gives,

$1.75x+1.75y=962.5\\\\\\1.75x+2.25y=1153\\1.75x+1.75y=962.5\\(-)\\-------------\\0.5y=190.5\\\\y=\frac{190.5}{0.5}=381$

So, there were 381 adults at the pool. Now, the number of children is equal to:

$x=550-y=550-381=169$

Therefore, there were 169 children at the pool.

6 0

3 years ago