Answer:


And using a calculator, excel ir the normal standard table we have that:

And we can calculate the probability like this:
Step-by-step explanation:
A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?
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 variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and 
Since the distribution for X is normal then we know that the distribution for the sample mean
is given by:

We can find the probability required like this:


And using a calculator, excel ir the normal standard table we have that:

And we can calculate the probability like this:

Answer:
Option A
Step-by-step explanation:
A type I error is committed when a researcher rejects the null hypothesis when it is actually true.
The null hypothesis is: U <= 50%
The alternative is: U > 50%
Thus, the principal could have committed an error by rejecting null hypothesis and concluding that more than 50% of students want earlier lunch, when in actuality 50% or less want earlier lunch.
Given:


To find:
The rate of change in volume at 
Solution:
We know that, volume of a cone is

Differentiate with respect to t.
![\dfrac{dV}{dt}=\dfrac{1}{3}\pi\times \left[(r^2\dfrac{dh}{dt}) + h(2r\dfrac{dr}{dt})\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%5Ctimes%20%5Cleft%5B%28r%5E2%5Cdfrac%7Bdh%7D%7Bdt%7D%29%20%2B%20h%282r%5Cdfrac%7Bdr%7D%7Bdt%7D%29%5Cright%5D)
Substitute the given values.
![\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cdfrac%7B22%7D%7B7%7D%5Ctimes%20%5Cleft%5B%28120%29%5E2%28-2.1%29%20%2B175%282%29%28120%29%281.4%29%5Cright%5D)
![\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B22%7D%7B21%7D%5Ctimes%20%5Cleft%5B-30240%2B58800%5Cright%5D)


Therefore, the volume of decreased by 29920 cubic inches per second.
Answer:
Equal df
Step-by-step explanation:
Given that a chi square test for goodness of fit is used to examine the distribution of individuals across three categories,
Hence degree of freedom = 3-1 =2
Similarly for a chi-square test for independence is used to examine the distribution of individuals in a 2×3 matrix of categories.
Here degree of freedom = (r-1)(c-1) where r = no of rows and c =no of columns
= (2-1)(3-1) = 2
Thus we find both have equal degrees of freedom.