Statistics Lecture 6.3: The Standard Normal Distribution. Using z-score, Standard Score

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[@taekwondotime]

*Advanced: Using the Normal Distribution to Approximate the Binomial Distribution* -- Students should be aware that the normal distribution has a lot in common with the binomial distribution. In fact, if np >=5 and nq >= 5, then you can use the normal distribution instead of using the binomial distribution, and the accuracy of the result will be good. Often times, you have no choice in the matter. Consider the following problem: Example: "If 20% of people prefer a white phone over any other color available, what is the probability that among the next 1000 phones purchased, between 170 and 185 inclusive will be white?" If you recall from chapter 5, you would have n = 1000, p = 0.20, and q = 1 - p = 0.80. At this point, students would flip to their binomial table and (much to their chagrin) would realize that the table is of no help to them, because it doesn't go up to values of n = 1000. So the only way you could solve this problem (using a binomial distribution) would be to use the formula to calculate the probability for each value between 170 and 185 and then add all of them up. As the numbers get larger, this quickly becomes nearly impossible to do. Here's where the magic happens... We use the following Z score formula: *z = (x - np) / sqrt(npq)* because... mean = np std. dev. = sqrt(npq) We also extend the range like we did before to accommodate the rounded values (shifting discrete data to continuous data). Therefore.... P(170 Area = 0.1251 Subtracting the areas we get: 0.1251 - 0.0080 = 0.1171. Therefore, there is a 11.71% chance that of the next 1000 people selected at random, between 170 and 185 will choose a white phone.

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iLoveTurtlesHaha is my answers true?

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Is my answer in ex no.2 is 5.5 true.

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Is zscore =10.1 true

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Students minds get blown @41:30 lol. Up until this point they thought car manufacturers pulled numbers out of their a** when designing car seats. The irony is that 99.5% of these students will never use the knowledge they gain in this class because they still don't "get it". They lack the critical thinking aspect that goes with stats. To them it's just formulas and weird symbols, and they forget most of that the moment they walk out of the classroom from the sound of things. Statistics/Probability theory is quite possibly the most powerful area in mathematics. If you know stats well, you pretty much rule the world. The knowledge is so widely applicable in all jobs and in most parts of everyday life that it's difficult to find something that isn't affected by statistics.

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[@0AlHidaya0]

Hey professor, please I have question about normal distribution when we need to calculate the peobability of "At least one". this is the exercise and it's about question "e": Scores on an achievement test are known to be normally distributed with a mean of 420 and a standard deviation of 80. a. For a randomly chosen person taking this test, what is the probability of a score between 400 and 480? b. What is the minimum test score needed in order to be in the top 10% of all people taking the test? c. For a randomly chosen individual, state, without doing the calculations, in which of the following ranges his score is most likely to be: 400-439, 440-479, 480-519, or 520-559. d. In which of the ranges listed in part (c) is the individual’s score least likely to be? e. Two people taking the test are chosen at random. What is the probability that at least one of them scores more than 500 points? Thanks for your help

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[@taekwondotime]

Advanced: Discrete Distribution Approximation -- Students should be aware that if the units they're dealing with are "rounded to the nearest something", they need to adjust the upper and lower bounds when calculating z-values. Here's an example: The tensile strength of a certain metal compound is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter. What proportion of these components exceed 10,150 kilograms per square centimeter in tensile strength? What makes this question different from the others is the fact that the units are rounded to the nearest 50 kg/cm^2 . If you tried to calculate the z score using z = (10150 - 10000) / 100 = 1.50, your answer would be wrong! To solve this type of problem you need to divide the rounded off measurement by two. Therefore, 50 / 2 = 25. You then add 25 to your upper bound: 10150 + 25 = 10175. Calculate your z score as follows: z = (10175 - 10000) / 100 = 1.75 When you look up the area you get: 0.9599. Therefore, P(X > 10150) to the nearest 50kg/cm^2 = P(X > 10175) = P(Z > 1.75) = 1 - P(Z < 1.75) = 1 - 0.9599 = 0.0401. Therefore, 4.01% of the components will exceed 10,150 kg/cm^2 in tensile strength. Why it's different: In order to take into account all of the values between 10150 and 10175 that were rounded down to 10150, the upper bound is shifted higher to include them.

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13:51 great I just ripped up my text book and I wasn’t even the one who messed up 😂

[@ahmadfadda2654]

In the last example shouldn't the Z-score be -2.57 not -1.645?

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Now I’m understanding,can you do another class on Z score

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[@hanhtooaein6368]

Can anyone explain the reason why Professor took shortest 5% instead of shortest 95% in 46:18? What would happen in real-life situation if we took shortest 95%?

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6.3 lesson is not working anymore~ Can you fix it, please??? You're teaching me better than my teacher could and English is not even my native language.

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[@thelthrythquezada8397]

what if I dont have a calculator like that right now? is there a way I can get the .9525, .1587 numbers?

[@victortsado9180]

The lecture is well understood , but i have issue with the reading of the Z table corresponding with the values you are getting , please i what you to share more light on it prof.

[@Glen_13]

how is the area found cause I just hear the lecturer saying check on the table

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[@christianhorta1945]

The calculator method is not working on my TI-84 plus. HELP!!!

[@ViperLemony]

yeah me too.. i was trying to do the -1 and then the 1.64 and then it came another number which was different from the table values.. have u figured it out?

[@christianhorta1945]

+chivy d i have no clue 😅 i managed to barely pass the class

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and he was your professor? hes awesome no? yeah i figured it out on the calculator so that is good at least.. i was doing the wrong function so yeah

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chivy d

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9:53 , is the answer is equals less than 174 lbs (OR) less than or equal to 174 lbs ?

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Thanks, pro!!!

[@alexrodriguez1008]

for the z formula= i didvided 174-172/29= 0.0689 wouldnt it be 0.069 since we have to round to 2 decimal places ? or is each instructor different at rounding. BTW your videos are dope! thanks

[@BalkanManic]

Two decimal places would be at .06

[@Abss0966]

8:52 why -10 ? I dont understand

[@user-ru6nn1wu4b]

Can anyone explain where he gets 2.575? It's at about 36:53.

[@nauroznausherwani769]

Good Question : In previous lectures professor explained how to translate percentage or probabilities into Z-score without involving X. Look in the table for probability value of 0.995( or close to that value) then add corresponding row z value and corresponding column z value you will get the answer.

[@sanzidaakter7710]

How did they get 0.5279?

[@Peter_1986]

The value 0.5279 tells you the proportion of the left part of the normal distribution curve, and this proportion is given at the point where z = 0.07; remember, the value z = 0 is at the exact center of the curve, so you can intuitively see that a z-value of 0.07 is slightly more than half of that curve (in other words, 0.5279). You can find this proportion either by looking up the z-score 0.07 in a table, or by using the "NormCD" function on calculators, and assuming that the leftmost value is maybe -10 or so; this is _very_ far to the left on the normal distribution curve, since the area under the curve becomes negligibly small at that point, and thus it can pretty much be considered negative infinity, and this ensures that you get almost the whole left part of the normal distribution curve. This would be written as "NormCD(-10, 0.07)", which should give the proportion 0.5279. Another interesting way to get the value 0.5279 is by actually using integrals; the integral for the area of a normal distribution curve is given by ∫ [ exp(-x²/2) / √(2⋅π) ]⋅dx, where you would again plug in the values -10 and 0.07 in order to get the proportion 0.5279. The NormCD function is a more compact version of that integral.

[@alicezvinoira3711]

@36:48 HELP HOW DID WE GET 2.575 PLEASE HELP

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Just a comment with correct spelling and grammar passing through. Don't mind me.

[@mrsomera2293]

Is 11.5 true

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BLESS U

[@shirleys2889]

1.67 = .4525

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superman :)

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I am terrible at this.

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