Answer:
Step-by-step explanation:
N⁶ · n⁵ · n⁴ ÷ n³ · n² ÷ n
n¹¹ · n⁴ ÷ n³ · n² ÷ n
n¹⁵ ÷ n³ · n² ÷ n
n⁵ · n² ÷ n
n¹⁰ ÷ n
n⁹
Answer:
3.55-3.59
Step-by-step explanation:
to round off to the nearest whole number, look at the first number after the decimal, if it is less than 5, add zero to the units term, If it is equal or greater than 5, add 1 to the units term.
So numbers that would be rounded off to 3.6 have to have an hundredth value equal or greater than 5
they include
3.55
3.56
3.57
3.58
3.59
Answer:
-13 is the answer I think so but wait for others also because my could be wrong
Answer:
![P(9.6 < \bar X](https://tex.z-dn.net/?f=%20P%289.6%20%3C%20%5Cbar%20X%20%3C15.6%29%20%3DP%28-0.8%3Cz%3C1.2%29%3D%20P%28Z%3C1.2%29-%20P%28Z%3C-0.8%29)
And using the normal standard distribution or excel we got:
![P(9.6 < \bar X](https://tex.z-dn.net/?f=%20P%289.6%20%3C%20%5Cbar%20X%20%3C15.6%29%20%3DP%28-0.8%3Cz%3C1.2%29%3D%20P%28Z%3C1.2%29-%20P%28Z%3C-0.8%29%3D0.8849-%200.2119%3D0.6731%20)
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".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
Where
and
Since the dsitribution for x is normal then we know that the distribution for the sample mean
is given by:
We want to find this probability:
![P(9.6 < \bar X](https://tex.z-dn.net/?f=%20P%289.6%20%3C%20%5Cbar%20X%20%3C15.6%29)
And we can use the z score formula given by;
![z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B%5Cbar%20X%20-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
And if we find the z score for the limits given we got:
![z = \frac{9.6-12}{\frac{6}{\sqrt{4}}}= -0.8](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B9.6-12%7D%7B%5Cfrac%7B6%7D%7B%5Csqrt%7B4%7D%7D%7D%3D%20-0.8)
![z = \frac{15.6-12}{\frac{6}{\sqrt{4}}}= 1.2](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B15.6-12%7D%7B%5Cfrac%7B6%7D%7B%5Csqrt%7B4%7D%7D%7D%3D%201.2)
So we can calculate this probability like this:
![P(9.6 < \bar X](https://tex.z-dn.net/?f=%20P%289.6%20%3C%20%5Cbar%20X%20%3C15.6%29%20%3DP%28-0.8%3Cz%3C1.2%29%3D%20P%28Z%3C1.2%29-%20P%28Z%3C-0.8%29)
And using the normal standard distribution or excel we got:
![P(9.6 < \bar X](https://tex.z-dn.net/?f=%20P%289.6%20%3C%20%5Cbar%20X%20%3C15.6%29%20%3DP%28-0.8%3Cz%3C1.2%29%3D%20P%28Z%3C1.2%29-%20P%28Z%3C-0.8%29%3D0.8849-%200.2119%3D0.6731%20)
25% of people leave early as 6 is 1/4 of 24. You can solve this by dividing 6 by 24 (I believe, maybe check with someone on that but that’s how i figured it out).