Hi there! If you are stuck on how he got the value of 1.28. Using excel or a TI-84 pull up the invNorm function and plug in the value of the Z-table you would like to find, for this example, it will be .9. In excel or the calculator, it will look like this invNorm(.9) and click enter! Hope this helps some people!
In-case anyone is still stuck on this despite the comments, what you can do instead is 0.5 - 0.1 which is 0.4 and read from the z-score table which is closest to that value, in this case its 0.3997 so it should still be 1.28 also
I've double checked the table. The percentage 0.8997 lies at the intersection of 1.20 and 0.08. For a minute I thought you might have been looking at the negative half of the table, where 0.3997 might be, but that percentage doesn't even exist on the table. Please make sure you are using a Cumulative Standard Normal Distribution table, with "z" at the top left.
A standard normal distribution has a mean of 60 and a standard deviation of 7, what is the z-score for a score of 69? What is the probability of obtaining a z-score this high or higher? Interpret the z-score of 69? Can you help? How would I figure that and what is my correct answer?
Can u help me with a problem? I have P[Z*S >= 600000] =0.99. In one of the steps the value for 0.01 is -2.3263( negative value). But the table gives a positive value. Why is the value negative? Thanks in advance!
I need more information to properly answer your question. You say "the value for 0.01 is -2.3263." I am assuming that -2.3263 is a z-score, but what does 0.01 represent? Is it a probability (percentage) or an actual data value?
lencage Thank you for getting back. I've posted the question below. The second part of the question is what i don't understand:- The annual yields from a particular fund are independent and identically distributed. Each year, the distribution of (1+i) is log-normal with parameters =0.07µ and 2 =0.006 σ , where i denotes the annual yield on the fund. (i) Find the mean accumulation in ten years time of an investment in the fund of £20,000 at the end of each of the next ten years, together with £150,000 invested immediately. [5] (ii) Find the single amount which should be invested in the fund immediately to give an accumulation of at least £600,000 in ten years time with probability 0.99. [7] [Total 12] I meant the value for 1% (1-0.99) in the tables is 2.3263. But it says -2.3263 in the answers.
lencage these are the steps:- P[S >= 600000/Z]=.99 Now, [ (Log S - 10µ)/ σ*sqrt of 10] Follows N(0,1) [ (Log{600000/Z} - 10µ)/ σ*sqrt of 10] =0.01 [ (Log{600000/Z} - 10µ)/ σ*sqrt of 10]= -2.3263. I'm so sorry that bothering you! I'd really appreciate it if you can help me out with this! Thank you.
Because the distribution is normal it is necessary to use the z distribution table. You can also use your calculator if it has the appropriate statistics functions. I can help you with that.
lencage ah thank you. There was something wrong with my calculator and I had my maths exam earlier today so I was looking up other ways of solving this. luckily I figured out how to over ride the problem.
For this problem you must find the percentage or probability closest to the value of 0.90 in the body of a normal distribution table (or z table). Because that exact value is not in the body of the table, you must use 0.8997. Looking all the way to the left of the table you find under the "z" column there is 1.20. Looking all the way to the top of the table you find 0.08. Adding the two values gives you 1.28, which is the corresponding z value for a percentage (probability) of 0.8997. Now that you have the z value you can use the formula X = z value*(std dev.) + mean to calculate the X value score at the 90th percentile (the minimum passing score on the exam). The method that I describe in this video is counterintuitive to most beginning statistics students because they are accustomed to using the normal distribution table to find a probability when first given a z score. The method in this video works the table in reverse order. You can estimate, based upon the mean score given as 200 with std dev of 20, and using what you know about the normal distribution that the z value for the 90th percentile must be somewhere between 1 and 2. In other words, the 90th percentile score has to be between 1 and 2 standard deviations above the mean.
Hi there! If you are stuck on how he got the value of 1.28. Using excel or a TI-84 pull up the invNorm function and plug in the value of the Z-table you would like to find, for this example, it will be .9. In excel or the calculator, it will look like this invNorm(.9) and click enter! Hope this helps some people!
I was stuck on an assignment question that was similar to this for weeks! Finally sorted it with your help! Thanks a bunch!
You're welcome! I'm glad I could help!
@@lencage Thank you!!!
In-case anyone is still stuck on this despite the comments, what you can do instead is 0.5 - 0.1 which is 0.4 and read from the z-score table which is closest to that value, in this case its 0.3997 so it should still be 1.28 also
Thank you
Just what I was looking for, thank you!!!!
Don't ever step into a classroom.
U do not take preparation or teaching seriously
omg chill. not everybody has unlimited time to spend hrs on a video.
Thank you so much! This helped me finished my assignment that's due today. Ah lifesaver🤧🤧💖💖
You didnt even share how to get the z value. what table are you looking at. this makes me more confused
Katina Pittman the standard normal distribution table
good video if you already have cursory knowledge in this area and need help only with this specific problem. thank you!
Hey how did you get the 1.28? bit confused about that..
from the table
@@asaad4349 could you please more specific？
@@sherryx4752 using Table A that’s corresponds to normal distribution
Thank you so much, I was able to solve a problem I was stuck on with this explanation!
where did 1.29 come from??
oh.. never mind.. i got it..just what i've been looking for.
thanks.
I've double checked the table. The percentage 0.8997 lies at the intersection of 1.20 and 0.08. For a minute I thought you might have been looking at the negative half of the table, where 0.3997 might be, but that percentage doesn't even exist on the table. Please make sure you are using a Cumulative Standard Normal Distribution table, with "z" at the top left.
A standard normal distribution has a mean of 60 and a standard deviation of 7, what is the z-score for a score of 69? What is the probability of obtaining a z-score this high or higher? Interpret the z-score of 69? Can you help? How would I figure that and what is my correct answer?
THANK YOU, YOU ARE A LEGEND!!
Great video! Thank you!
to find Z
in google sheet
Z=norm.s.inv(0.9)=1.28
Thank you! I am going to repost so more people see this!
i still dont understand why 0.90 has been changed to 0.8997.
A z-score is negative when its corresponding X value is to the left of the mean, or mathematically, less than the mean.
Thank you I have an exam in a few hours and now know how to do this :-)
You're welcome. I'm glad that I was able to help you. How was your exam?
I got a B in the exam really thought I had failed.
Ciara mc It's good to know that you did well on the exam. Congratulations!
You're welcome. Please subscribe.
thank you!!!! ahhh its all soo simple now
@HBCult and @shanisdaman
You're welcome!
thanks for the help.
thank you so much i was worried I wasn't going to get points on my homework
Confusing at first with .9= 1.28 got it now! Thanks for the assist! hmm i was using the wrong table! :)
how.... the 2 numbers that intersect, but then what
To get 1.28, use the INVERSE cumulative distribution table. 0.9=1.2816
Terrible.
THANK YOU
I have a question. Can a survey measured in percent exceed 100?
Never mind, I paused too soon. Thank you!
all wrong
lencage, do you happen to know where you got these practice problems from?
how did he get 1-0.10?
This guy does everything in slow motion.....
I recorded this video while teaching an actual face-to-face class, which is why I took my time with the lesson. No good teacher rushes a lesson! :)
Can u help me with a problem? I have P[Z*S >= 600000] =0.99. In one of the steps the value for 0.01 is -2.3263( negative value). But the table gives a positive value. Why is the value negative? Thanks in advance!
I need more information to properly answer your question. You say "the value for 0.01 is -2.3263." I am assuming that -2.3263 is a z-score, but what does 0.01 represent? Is it a probability (percentage) or an actual data value?
All probabilities or percentages less than 50% (or 0.5) have corresponding z-scores that are negative.
lencage Thank you for getting back. I've posted the question below. The second part of the question is what i don't understand:-
The annual yields from a particular fund are independent and identically distributed. Each year, the distribution of (1+i) is log-normal with parameters =0.07µ and 2 =0.006 σ , where i denotes the annual yield on the fund.
(i) Find the mean accumulation in ten years time of an investment in the fund of £20,000 at the end of each of the next ten years, together with £150,000 invested immediately. [5]
(ii) Find the single amount which should be invested in the fund immediately to give an accumulation of at least £600,000 in ten years time with probability 0.99. [7] [Total 12]
I meant the value for 1% (1-0.99) in the tables is 2.3263. But it says -2.3263 in the answers.
lencage these are the steps:-
P[S >= 600000/Z]=.99 Now, [ (Log S - 10µ)/ σ*sqrt of 10] Follows N(0,1)
[ (Log{600000/Z} - 10µ)/ σ*sqrt of 10] =0.01
[ (Log{600000/Z} - 10µ)/ σ*sqrt of 10]= -2.3263.
I'm so sorry that bothering you! I'd really appreciate it if you can help me out with this! Thank you.
thanks alot. but when does the z value become negative??
Baller
I have no clue.
where did 1.20 and 0.08 come from?
why is there so many dislikes?
When i look up 0.90 in the percentage table I get a z value of 1.28... Where did 0.8159 come from?
could you explain how you get z value of 1.28 from table
You're welcome. Please subscribe. There will be more videos soon.
what if we don't use the Z table... I've never heard of it before but I understand everything else...
Because the distribution is normal it is necessary to use the z distribution table. You can also use your calculator if it has the appropriate statistics functions. I can help you with that.
lencage ah thank you. There was something wrong with my calculator and I had my maths exam earlier today so I was looking up other ways of solving this. luckily I figured out how to over ride the problem.
so confusing, why didn't you just do it over right?
When I did this video I did not have time to redo the lesson. I apologize for any confusion.
Thank you, very helpful
1.28*
For this problem you must find the percentage or probability closest to the value of 0.90 in the body of a normal distribution table (or z table). Because that exact value is not in the body of the table, you must use 0.8997. Looking all the way to the left of the table you find under the "z" column there is 1.20. Looking all the way to the top of the table you find 0.08. Adding the two values gives you 1.28, which is the corresponding z value for a percentage (probability) of 0.8997. Now that you have the z value you can use the formula X = z value*(std dev.) + mean to calculate the X value score at the 90th percentile (the minimum passing score on the exam). The method that I describe in this video is counterintuitive to most beginning statistics students because they are accustomed to using the normal distribution table to find a probability when first given a z score. The method in this video works the table in reverse order. You can estimate, based upon the mean score given as 200 with std dev of 20, and using what you know about the normal distribution that the z value for the 90th percentile must be somewhere between 1 and 2. In other words, the 90th percentile score has to be between 1 and 2 standard deviations above the mean.
you need to know what you doing before you get on here trying to teach something, im completely lost now thanks
Zak McKellar like I said. You can shut up I’m not the only person in the comments that feels he don’t know what he talking about which he doesn’t!
the closest is not 0.8997 but 0.9015...
60 people do not like math
the closest is not 0.8997 but 0.9015...
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