The probability that the number 72 bus is on time is

Mathematics

2 answers:

Colt1911 [192]4 years ago

6 0

Answer:

<h2>Probability that it will be late = 1-0.76 = 0.24</h2>

<h3>Hope it helps........</h3>

Levart [38]4 years ago

4 0

0.24 is the answer to this question

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A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.

AVprozaik [17]

Answer:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

8 0

3 years ago

How would you solve for side a? Pythagorean theorem, cause if that's the case wouldn't

Yuri [45]

The angle is 127°, so the Pythagorean theorem does not apply.

Generally, you would use the Law of Cosines to solve for the third side when given two sides and the angle between them.

EF^2 = 7^2 +15^2 -2*7*15*cos(127°)
EF ≈ √(400.3812) ≈ 20.00953

My guess is that you're expected to use 20 for the length of the 3rd side.

By Heron's formula, s = (7 +15 +20)/2 = 21

Area = √(21*(21 -7)*(21 -15)*(21 -20)) = √(21*14*6*1) = √1764 = 42 . . . . yd^2

_____
More directly,
.. Area = (1/2)*7*15*sin(127°) ≈ 41.928 . . . . yd^2