Answer:
(a) The probability of the event (<em>X</em> > 84) is 0.007.
(b) The probability of the event (<em>X</em> < 64) is 0.483.
Step-by-step explanation:
The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.
The probability mass function of a Poisson distribution is:
(a)
Compute the probability of the event (<em>X</em> > 84) as follows:
P (X > 84) = 1 - P (X ≤ 84)
![=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007](https://tex.z-dn.net/?f=%3D1-%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D84%7D%5Cfrac%7Be%5E%7B-64%7D%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3D1-%5Be%5E%7B-64%7D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D84%7D%5Cfrac%7B%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5D%5C%5C%3D1-%5Be%5E%7B-64%7D%5B%5Cfrac%7B%2864%29%5E%7B0%7D%7D%7B0%21%7D%2B%5Cfrac%7B%2864%29%5E%7B1%7D%7D%7B1%21%7D%2B%5Cfrac%7B%2864%29%5E%7B2%7D%7D%7B2%21%7D%2B...%2B%5Cfrac%7B%2864%29%5E%7B84%7D%7D%7B84%21%7D%5D%5D%5C%5C%3D1-0.99308%5C%5C%3D0.00692%5C%5C%5Capprox0.007)
Thus, the probability of the event (<em>X</em> > 84) is 0.007.
(b)
Compute the probability of the event (<em>X</em> < 64) as follows:
P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)
![=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483](https://tex.z-dn.net/?f=%3D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D63%7D%5Cfrac%7Be%5E%7B-64%7D%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3De%5E%7B-64%7D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D63%7D%5Cfrac%7B%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3De%5E%7B-64%7D%5B%5Cfrac%7B%2864%29%5E%7B0%7D%7D%7B0%21%7D%2B%5Cfrac%7B%2864%29%5E%7B1%7D%7D%7B1%21%7D%2B%5Cfrac%7B%2864%29%5E%7B2%7D%7D%7B2%21%7D%2B...%2B%5Cfrac%7B%2864%29%5E%7B63%7D%7D%7B63%21%7D%5D%5C%5C%3D0.48338%5C%5C%5Capprox0.483)
Thus, the probability of the event (<em>X</em> < 64) is 0.483.
Answer:
the mean and standard error of the mean are 200 and 2 respectively.
Step-by-step explanation:
Given that ;
the sample size n = 81
population mean μ = 200
standard deviation of the infinite population σ = 18
A population is the whole set of values, or individuals you are interested in, from an experimental study.
The value of population characteristics such as the Population mean (μ), standard deviation (σ) are said to be known as the population distribution.
From the given information above;
The sample size is large and hence based on the central limit theorem the mean of all the means is same as the population mean 200.
i.e
= 200
∴ The mean = 200
and the standard error of the mean can be determined via the relation:
Therefore ; the mean and standard error of the mean are 200 and 2 respectively.
Answer:
The bush after the two weeks was 14.4 inches tall
Step-by-step explanation:
Let
x ----> bush height after two weeks
we know that
The bush height after two weeks is equal to multiply the height when it was first planted by 1.20
so
Answer:
1290 jumps
Step-by-step explanation:
86 jumps x 15 min = 1290 jumps
Can you make the answer clearer. I would be glad to help you. I just don't understand where it says "the sum of three numbers is 16 the sum of twice the first number"
