Answer:
Step-by-step explanation:
hello :
x ÷ y for x = 8 and y = 16 means : 8/16 = 8/(2×8) = 1/2
Answer:
564 pencils
Step-by-step explanation:
1 box = 12 pencils
25 boxes = 25 × 12 pencils = 300 pencils
22 boxes = 22 × 12 pencils = 264 pencils
300 + 264 = 564 pencils
Answer:
The answer is 32.
Step-by-step explanation:
Answer:
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Step-by-step explanation:
Previous concepts
The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.
![IQR= Q_3 -Q_1](https://tex.z-dn.net/?f=IQR%3D%20Q_3%20-Q_1)
The standard deviation is a measure of dispersion obatined from the sample variance and is given by:
![s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%3D%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
Solution to the problem
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Answer:
The correct answer is "0.0000039110".
Step-by-step explanation:
The given values are:
![Y_n\rightarrow N(\mu, \sigma^2)](https://tex.z-dn.net/?f=Y_n%5Crightarrow%20N%28%5Cmu%2C%20%5Csigma%5E2%29)
![\mu = 40n](https://tex.z-dn.net/?f=%5Cmu%20%3D%2040n)
![\sigma^2=100n](https://tex.z-dn.net/?f=%5Csigma%5E2%3D100n)
![n=20](https://tex.z-dn.net/?f=n%3D20)
then,
The required probability will be:
= ![P(Y_{20}>1000)](https://tex.z-dn.net/?f=P%28Y_%7B20%7D%3E1000%29)
= ![P(\frac{Y_{20}-\mu}{\sigma} >\frac{1000-40\times 20}{\sqrt{100\times 20} } )](https://tex.z-dn.net/?f=P%28%5Cfrac%7BY_%7B20%7D-%5Cmu%7D%7B%5Csigma%7D%20%3E%5Cfrac%7B1000-40%5Ctimes%2020%7D%7B%5Csqrt%7B100%5Ctimes%2020%7D%20%7D%20%29)
= ![P(Z>\frac{1000-800}{44.7214} )](https://tex.z-dn.net/?f=P%28Z%3E%5Cfrac%7B1000-800%7D%7B44.7214%7D%20%29)
= ![P(Z>\frac{200}{44.7214} )](https://tex.z-dn.net/?f=P%28Z%3E%5Cfrac%7B200%7D%7B44.7214%7D%20%29)
= ![P(Z>4.47)](https://tex.z-dn.net/?f=P%28Z%3E4.47%29)
By using the table, we get
= ![0.0000039110](https://tex.z-dn.net/?f=0.0000039110)