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Anni [7]
3 years ago
6

Dawn and Pablo are members of a normally distributed population that is being sampled. If the chance of Dawn being included in t

he sample is 0.05%, what must the chance of Pablo being included in the sample be in order to be able to make inferences about the population based on the sample?
A) 0%
B) greater than 0% and less than 0.05%
C) 0.05%
D) greater than 0.05%
Mathematics
2 answers:
777dan777 [17]3 years ago
7 0
Answer: option C) 0.05%

In a normal distribution there are no biases, the chances (probabilities) of any member to be selected are the same. So, the chance to choose Pablo are the same that the chance to choose Dawn.
lys-0071 [83]3 years ago
7 0

Answer:

C. 0.05%

Step-by-step explanation:

NONE

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Please help. Which number(s) below belong to the solution set of the inequality? Check all that apply. 9x > 117
lord [1]

Answer:

A, d and e

Step-by-step explanation:

So take the numbers a multiply them by nine. if they are more than 117, then they are correct, but with D 9x13 is equal to 117 so it is part of the solution set.

8 0
3 years ago
estaba previsto destinar 3/14 de partes de una finca a plazas de aparcamiento pero finalmente han destinado 3/4 de lo previsto a
JulsSmile [24]

Queremos ver que fracción de la finca se ha destinado a plazas de aparcamiento.

La solución es:

La fracción de la finca que se destina a plaas de aparcamiento es 3/56

Sabemos que originalmente se iba a destinar 3/14 del total de la finca a plazas de aparcamiento, pero finalmente se destino 3/4 de lo previsto a zonas ajardinadas.

Es decir, <u>se destino 3/4 de los 3/14 del total de la finca</u> a zonas ajardinadas, entonces <u>el 1/4 restante se dedico a plazas de aparcamiento</u>, esto da:

(1/4)*3/14  = 3/56

La fracción de la finca que se destina a plaas de aparcamiento es 3/56

Sí quieres aprender más, puedes leer:

brainly.com/question/16649102

8 0
3 years ago
Workout 2/5 of 16 miles
yaroslaw [1]

Answer:

6.4

Step-by-step explanation:

16 ÷ 2/5 = 6.4!

that's it

3 0
3 years ago
Geometry help will give brainliest
Natali [406]
6x + 4x + 10 = 180
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Answer:
mCA = 78°
8 0
3 years ago
The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand
Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 18.6, \sigma = 5.9

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 18.6}{5.4}

Z = 0.44

Z = 0.44 has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{20.4 - 18.6}{0.6768}

Z = 2.66

Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

4 0
3 years ago
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