Please see the item below. I need help with part 4. The answers shown came back as being incorrect. Please help! You can complete this problem using either Minitab or Geogebra, but you will likely find the visualization of Geogebra easier to understand. Instructionsare given here for Geogebra, but you can review the video “Using Minitab with the Binomial Distribution" in D2L under Content>StatisticsResources>Minitab Tutorials by Dr. Matos to see how to complete the problem using Minitab.Open Geogebra and choose "Probability Calculator." You can also get to this through the "View" menu. This opens in a new window (the online/appversion may behave differently). The default is a Normal distribution with mean 0 and standard deviation 1 (the Standard Normal).Select Normal from the drop-down menu just below the graph, and choose "Binomial." The default is n = 20 observations/trials withp 0.5 probability ofsuccess on each trial. From 8 - 12 success are highlighted, with total probability 0.7368. You might have to resize the bottom part of the display to see%3Dthis. To the right of the graph, the probabilities for exactly k successes are shown.Click here for a screenshot.>Part 1:Part 2:> Part 3:- Part 4:Now changen to 200 and p to 0.85; this is the 85% passing example with 200 observations. Note that the probability of exactly k outcomes is oftenlisted as zero (in reality, they are very small positive numbers). Compute the probabilities of the following. Use four decimal places, and do not convertto percent.Exactly 170 successes:.0789Help me!• From 165 to 175 successes:0.6779Help me!• From 160 to 180 successes:0.9523Help me!....Between 160 and 180 successes:0.9523Help me!.... ...Fewer than 150 successes:0.00002Help me!Note that the graph looks like a Normal distribution. This is extremely important later!

Answered: Now change n to 200 and p to 0.85; this…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Please see the item below. I need help with part 4. The answers shown came back as being incorrect. Please help!

You can complete this problem using either Minitab or Geogebra, but you will likely find the visualization of Geogebra easier to understand. Instructions
are given here for Geogebra, but you can review the video “Using Minitab with the Binomial Distribution" in D2L under Content>Statistics
Resources>Minitab Tutorials by Dr. Matos to see how to complete the problem using Minitab.
Open Geogebra and choose "Probability Calculator." You can also get to this through the "View" menu. This opens in a new window (the online/app
version may behave differently). The default is a Normal distribution with mean 0 and standard deviation 1 (the Standard Normal).
Select Normal from the drop-down menu just below the graph, and choose "Binomial." The default is n = 20 observations/trials withp 0.5 probability of
success on each trial. From 8 - 12 success are highlighted, with total probability 0.7368. You might have to resize the bottom part of the display to see
%3D
this. To the right of the graph, the probabilities for exactly k successes are shown.
Click here for a screenshot.
>Part 1:
Part 2:
> Part 3:
- Part 4:
Now changen to 200 and p to 0.85; this is the 85% passing example with 200 observations. Note that the probability of exactly k outcomes is often
listed as zero (in reality, they are very small positive numbers). Compute the probabilities of the following. Use four decimal places, and do not convert
to percent.
Exactly 170 successes:
.0789
Help me!
• From 165 to 175 successes:
0.6779
Help me!
• From 160 to 180 successes:
0.9523
Help me!
....
Between 160 and 180 successes:
0.9523
Help me!
.... ...
Fewer than 150 successes:
0.00002
Help me!
Note that the graph looks like a Normal distribution. This is extremely important later!

Expert Solution

Step 1

Binomial distribution is used to estimate the probability of x number of successes out of total n number of trials. A random variable follow the binomial distribution if there are only two possible outcomes of each trial and each trial is independent with equal probability of success.

The probability mass function of binomial distribution is $P (x) = (\begin{matrix} n \\ x \end{matrix}) p^{x} {(1 - p)}^{n - x}$ . In this probability mass function n is total number of trials, x is number of success and p is the probability of success. The mean number of successes in binomial distribution is $μ = n p$ .