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2 years ago

How can the probability of an event help make predictions

1 answer:

6 0

The probability of an event can help make predictions by knowing the chance of something happening and using your knowledge to determine the probability of choosing for example blue on a spinner out of 6 with 4 blues. there is 4 blues and 2 red so you predict it will be on red because there is more of it

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List below are the speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, CA. Use the sample data to construct

zheka24 [161]

Answer:

The 99% confidence interval of the population standard deviation is 1.7047 < σ < 7.485

Step-by-step explanation:

Confidence interval of standard deviation is given as follows;

$\sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}$

s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$

Where:

$\bar x$ = Sample mean

s = Sample standard deviation

n = Sample size = 7

χ = Chi squared value at the given confidence level

$\bar x$ = ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857

The sample standard deviation s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$ = 2.854

The test statistic, derived through computation, = ±3.707

Which gives;

$C. I. = 57.857 \pm 3.707 \times \dfrac{2.854}{\sqrt{7} }$

$\sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{16.812}^{}}}< \sigma < \sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{0.872}}$

1.7047 < σ < 7.485

The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.

7 0

2 years ago

A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib

Jet001 [13]

Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 502, \sigma = 115$

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:

X = 590:

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

$Z = \frac{590 - 502}{115}$

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

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

$Z = \frac{400 - 502}{115}$

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.7764 - 0.1867 = 0.5897 = 58.97%.

58.97% of students would be expected to score between 400 and 590.

More can be learned about the normal distribution at brainly.com/question/27643290

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6 0

1 year ago

Relate Fractions, Decimals, and Money. Write the money amount as a fraction in terms of dollars. The problems are $0.66 and $1.0

ozzi

You convert a decimal to a fraction by putting the digits to the right of the decimal point over "1" followed by the same number of zeros. (Most folks learn this in 3rd or 4th grade.)

0.66 = 66/100 . . . . . 2 decimals, so the denominator is 100

Then you reduce the fraction by canceling common factors. It helps to know your times tables and divisibility rules. (Or you can have your calculator do it.)

0.66 = 66/100 = 33/50
1.00 = 100/100 = 1

7 0

2 years ago

the total cost, c, of an item purchased from a store is c = p .07p, where p is the price before sales tax, and there is a 7% sal

alekssr [168]

C = p + 0.07p
c = p(1 + 0.07)
c = 1.07p

6 0

2 years ago

Find the percentage of increase from 80 to 90

antoniya [11.8K]

Answer: The answer is 1/2 1/4 and 1/9

Step-by-step explanation: I did it already and i got it right

5 0

2 years ago