Answer:
t(c) = 3.355
Step-by-step explanation:
We assume a normal distribution, and with sample size n = 9 we should follow a t -student test on both tails since the FDA is interested in determining if the amount of drug absorbed is different from 3.5 micrograms.
Therefore if α = 0,01 that means that confidence interval is 99 % or 0,99
Finally with α/2 = 0,005 and 8 degrees of freedom we find in t-student table t(c) = 3.355
Answer:
0.1587
Step-by-step explanation:
Let X be the commuting time for the student. We know that 
. Then, the normal probability density function for the random variable X is given by
![f(x) = \frac{1}{\sqrt{2\pi(5)^{2}}}\exp[-\frac{(x-\mu)^{2}}{2(5)^{2}}]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%5Cpi%285%29%5E%7B2%7D%7D%7D%5Cexp%5B-%5Cfrac%7B%28x-%5Cmu%29%5E%7B2%7D%7D%7B2%285%29%5E%7B2%7D%7D%5D)
. We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,
= 0.1587
To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)
The mean is 7.5 you add up all the numbers and then divide by the amount of numbers there is
1.) 7.2 2.) 15 3.)12
4.) 23 5.) 100, 100, 1000
Answer:
Step-by-step explanation:
Similar shapes also maintain a constant proportion of sides. Therefore, we can set up the following equation:
Solving, we get:
