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densk [106]
1 year ago
14

Confidence intervals for the means provide an estimate for where the true mean lies.

Mathematics
1 answer:
stepladder [879]1 year ago
3 0

It is true that the confidence intervals for the mean provide an estimate for where the true mean lies.

In statistics, a confidence interval denotes the likelihood that a population parameter will fall between a set of values for a given proportion of the time. A confidence interval depicts the likelihood that a parameter will fall between two values near the mean. Confidence intervals quantify the degree of uncertainty or certainty in a sampling procedure.

The mean is a basic mathematical average of two or more values. There are two sorts of means that may be calculated: the arithmetic mean and the geometric mean. A mean tells you the average of a bunch of values, which helps you contextualize each data point.

To learn more about confidence interval, visit :

brainly.com/question/24131141

#SPJ4

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A credit card company decides to study the frequency with which its cardholders charge for items from a certain change of retail
gregori [183]

Answer:

15.87% of the total number of cardholder would be expected to be charging 27 or more in the study.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 25 charged purchases and a standard distribution of 2

This means that \mu = 25, \sigma = 2

Proportion above 27

1 subtracted by the pvalue of Z when X = 27. So

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

Z = \frac{27 - 25}{2}

Z = 1

Z = 1 has a pvalue of 0.8413

1 - 0.8413 = 0.1587

Out of the total number of cardholders about how many would you expect are charging 27 or more in the study?

0.1587*100% = 15.87%

15.87% of the total number of cardholder would be expected to be charging 27 or more in the study.

6 0
2 years ago
Solve by substitution
Tomtit [17]
X= -46 and y = -40 I worked it out I might be wrong
8 0
3 years ago
Read 2 more answers
What is the equation of the line whose graph is parallel to the graph of y=3x-10
TEA [102]
Hello,

y=3x+b is parallele to y=3x-10
In order to determine b, we must know something else.

3 0
3 years ago
Need help please.........................
Archy [21]

Answer:

16t^{2} + 2t - 6

Step-by-step explanation:

(12t^{2}  + 5t -y) - (-4t^{2} + 3t -1)

simplify

12t^{2} + 5t -7 +4t^{2} -3t +1

combine like terms

16t^{2} + 2t - 6

8 0
2 years ago
A cleaning company charges x dollars per hour to clean floors and y dollars per hour to clean the rest of the house
Nina [5.8K]

Answer:

(15,18)

Step-by-step explanation:

write two equations with the information provided.

2x + 3y = 84

x + 4y = 87

use the property of substitution to answer.

x + 4y = 87,   x = 87 - 4y

2(87 - 4y) + 3y = 84

174 - 8y + 3y = 84

174 - 84 = 8y - 3y

90 = 5y

90/5 = y

y = 18

Add the value of Y to an original equation. Solve for X

2x + 3(18) = 84

2x + 54 = 84

2x = 84 - 54

2x = 30

x = 30/2

x = 15

8 0
3 years ago
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