## Wednesday, May 6, 2020

### Normal Distribution Free Essays

Finda)P(Z gt; 2. 58),b)P(Z lt; -1),c)P(-1. 5 ( Z lt; 5) Ans : a) P( Z gt; 2. We will write a custom essay sample on Normal Distribution or any similar topic only for you Order Now 58) = 0. 0049 ( 4 decimal places) b) P ( Z lt; -1) = 0. 1587 ( 4 decimal places) c) P ( -1. 5? Z lt; 5) = P ( -1. 5 lt; Z lt; 5) = (0. 5- 0. 0668) + ( 0. 5 -0) = 0. 9332 ( 4 decimal places) 2. Find the value of z if the area under a Standard Normal curve a)to the right of z is 0. 3632; b)to the left of z is 0. 1131; c)between 0 and z, with z gt; 0, is 0. 4838; d)between -z and z, with z gt; 0, is 0. 9500. Ans : a) z = + 0. 35 ( find 0. 5- 0. 3632 = 0. 1368 in the normal table) b) z = -1. 1 ( find 0. 5 Ã¢â‚¬â€œ 0. 1131 = 0. 3869 in the normal table) c ) the area between 0 to z is 0. 4838, z = 2. 14 d) the area to the right of +z = ( 1-0. 95)/2 = 0. 025, therefore z = 1. 96 3. Given the Normally distributed variable X with mean 18 and standard deviation 2. 5, find a)P(X lt; 15); b)the value of k such that P(X lt; k) = 0. 2236; c)the value of k such that P(X gt; k) = 0. 1814; d)P( 17 lt; X lt; 21). Ans : X ~ N ( 18, 2. 52) a) P ( X lt; 15) P ( Z lt; (15-18)/2. 5) = P ( Z lt; -1. 2) = 0. 1151 ( 4 decimal places) b) P ( X lt; k) = 0. 2236 P ( Z lt; ( k Ã¢â‚¬â€œ 18) / 2. ) = 0. 2236 From normal table, 0. 2236 = -0. 76 (k-18)/2. 5 = Ã¢â‚¬â€œ 0. 76, solve k = 16. 1 c) P (X gt; k) = 0. 1814 P ( Z gt; (k-18)/2. 5 ) = 0. 1814 From normal table, 0. 1814 = 0. 91 (k-18)/ 2. 5 = 0. 91, solve k = 20. 275 d) P ( 17 lt; X lt; 21) P ( (17 -18)/2. 5 lt; Z lt; ( 21-18)/2. 5) P ( -0. 4 lt; Z lt; 1. 2) = 0. 8849 Ã¢â‚¬â€œ 0. 3446 = 0. 5403 ( 4 decimal places) 4. In a sample of 25 observations from a Normal Distribution with mean 98. 6 and standard deviation 17. 2, find: Ans: a) n = 25, [pic] = ( = 98. 6, [pic] = /n = 17. 2/(25 = 3. 44 [pic]( N (98. 6, 3. 442 ) Prob = P(92 lt; [pic] lt; 102) = P(92 lt; [pic] lt; 102) = P( (92-98. 6)/3. 44 lt; Z lt; (102-98. 6)/3. 44 ) = P( -1. 92 lt; Z lt; 0. 99) = 0. 8116 b)the corresponding probability given a sample of 36. Ans: b) n = 36, [pic] = ( = 98. 6, [pic] = /n = 17. 2/(36 = 2. 87 [pic]( N (98. 6, 2. 872 ) Prob = P(92 lt; [pic] lt; 102) = P(92 lt; [pic] lt; 102) = P( (92-98. 6)/2. 87 lt; Z lt; (102-98. 6)/2. 87 ) = P( -2. 30 lt; Z lt; 1. 19) = 0. 8723 ( * or 0. 8703 if rounding for 2. 87) 5. An X-ray technician is taking reading from her machine to ensure that it adheres to federal safety guidelines. She knows that the standard deviation of the amount of radiation emitted by the machine is 150 units, but she wants to take readings until the standard error of the sampling distribution is no higher than 25 units. How many reading should she take? Ans: Let X be the readings of radiation emitted by the machine. [pic] = /n = 150(n we want to find n such that 150/(n = [pic] ( 25 ( (n ( 6 ( n ( 36 She should take at least 36 readings. 6. Peter, researcher for a coffee corporation, is interested in determining the rate of coffee usage per household. He believes that yearly consumption per household is normally distributed with an unknown mean and a standard deviation of about 1. 25 pounds. a)If Peter takes a sample of 36 households and records their consumption of coffee for 1 year, what is the probability that the sample mean is within one-half pound of the population mean? Ans: Let X be the yearly consumption of coffee per household. n = 36, [pic]=( = ? , [pic] = /n = 1. 25(36 = 0. 2083 [pic]( N ([pic], 0. 20832 ) a)Prob = P(( Ã¢â‚¬â€œ 0. 5 lt; [pic] lt; ( + 0. 5) = P(- 0. 5/[pic] lt; Z lt; 0. /[pic]) = P( -0. 5/0. 2083 lt; Z lt; 0. 5/0. 2083 ) = P( -2. 4 lt; Z lt; 2. 4) = 0. 9836 b)How large a sample must be taken in order to be 98 percent certain that the sample mean is within one-half pound of the population mean? Ans: n = ? ,[pic] = /n = 1. 25/(n P(( Ã¢â‚¬â€œ 0. 5 lt; [pic] lt; ( + 0. 5) = 0. 98 ( P(-0. 5 lt; [pic]- ( lt; +0. 5) = 0. 98 ( P( -0. 5/1. 25/(n lt; Z lt; +0. 5/1. 25/(n ) = 0. 98 From Normal Table, we have P(-2. 33 lt; Z lt; 2. 33) = 0. 98 ( 0. 5/1. 25/(n = 2. 33 ( n = 33. 93 ( at least 34 households. How to cite Normal Distribution, Papers

### Normal Distribution Free Essays

Finda)P(Z gt; 2. 58),b)P(Z lt; -1),c)P(-1. 5 ( Z lt; 5) Ans : a) P( Z gt; 2. We will write a custom essay sample on Normal Distribution or any similar topic only for you Order Now 58) = 0. 0049 ( 4 decimal places) b) P ( Z lt; -1) = 0. 1587 ( 4 decimal places) c) P ( -1. 5? Z lt; 5) = P ( -1. 5 lt; Z lt; 5) = (0. 5- 0. 0668) + ( 0. 5 -0) = 0. 9332 ( 4 decimal places) 2. Find the value of z if the area under a Standard Normal curve a)to the right of z is 0. 3632; b)to the left of z is 0. 1131; c)between 0 and z, with z gt; 0, is 0. 4838; d)between -z and z, with z gt; 0, is 0. 9500. Ans : a) z = + 0. 35 ( find 0. 5- 0. 3632 = 0. 1368 in the normal table) b) z = -1. 1 ( find 0. 5 Ã¢â‚¬â€œ 0. 1131 = 0. 3869 in the normal table) c ) the area between 0 to z is 0. 4838, z = 2. 14 d) the area to the right of +z = ( 1-0. 95)/2 = 0. 025, therefore z = 1. 96 3. Given the Normally distributed variable X with mean 18 and standard deviation 2. 5, find a)P(X lt; 15); b)the value of k such that P(X lt; k) = 0. 2236; c)the value of k such that P(X gt; k) = 0. 1814; d)P( 17 lt; X lt; 21). Ans : X ~ N ( 18, 2. 52) a) P ( X lt; 15) P ( Z lt; (15-18)/2. 5) = P ( Z lt; -1. 2) = 0. 1151 ( 4 decimal places) b) P ( X lt; k) = 0. 2236 P ( Z lt; ( k Ã¢â‚¬â€œ 18) / 2. ) = 0. 2236 From normal table, 0. 2236 = -0. 76 (k-18)/2. 5 = Ã¢â‚¬â€œ 0. 76, solve k = 16. 1 c) P (X gt; k) = 0. 1814 P ( Z gt; (k-18)/2. 5 ) = 0. 1814 From normal table, 0. 1814 = 0. 91 (k-18)/ 2. 5 = 0. 91, solve k = 20. 275 d) P ( 17 lt; X lt; 21) P ( (17 -18)/2. 5 lt; Z lt; ( 21-18)/2. 5) P ( -0. 4 lt; Z lt; 1. 2) = 0. 8849 Ã¢â‚¬â€œ 0. 3446 = 0. 5403 ( 4 decimal places) 4. In a sample of 25 observations from a Normal Distribution with mean 98. 6 and standard deviation 17. 2, find: Ans: a) n = 25, [pic] = ( = 98. 6, [pic] = /n = 17. 2/(25 = 3. 44 [pic]( N (98. 6, 3. 442 ) Prob = P(92 lt; [pic] lt; 102) = P(92 lt; [pic] lt; 102) = P( (92-98. 6)/3. 44 lt; Z lt; (102-98. 6)/3. 44 ) = P( -1. 92 lt; Z lt; 0. 99) = 0. 8116 b)the corresponding probability given a sample of 36. Ans: b) n = 36, [pic] = ( = 98. 6, [pic] = /n = 17. 2/(36 = 2. 87 [pic]( N (98. 6, 2. 872 ) Prob = P(92 lt; [pic] lt; 102) = P(92 lt; [pic] lt; 102) = P( (92-98. 6)/2. 87 lt; Z lt; (102-98. 6)/2. 87 ) = P( -2. 30 lt; Z lt; 1. 19) = 0. 8723 ( * or 0. 8703 if rounding for 2. 87) 5. An X-ray technician is taking reading from her machine to ensure that it adheres to federal safety guidelines. She knows that the standard deviation of the amount of radiation emitted by the machine is 150 units, but she wants to take readings until the standard error of the sampling distribution is no higher than 25 units. How many reading should she take? Ans: Let X be the readings of radiation emitted by the machine. [pic] = /n = 150(n we want to find n such that 150/(n = [pic] ( 25 ( (n ( 6 ( n ( 36 She should take at least 36 readings. 6. Peter, researcher for a coffee corporation, is interested in determining the rate of coffee usage per household. He believes that yearly consumption per household is normally distributed with an unknown mean and a standard deviation of about 1. 25 pounds. a)If Peter takes a sample of 36 households and records their consumption of coffee for 1 year, what is the probability that the sample mean is within one-half pound of the population mean? Ans: Let X be the yearly consumption of coffee per household. n = 36, [pic]=( = ? , [pic] = /n = 1. 25(36 = 0. 2083 [pic]( N ([pic], 0. 20832 ) a)Prob = P(( Ã¢â‚¬â€œ 0. 5 lt; [pic] lt; ( + 0. 5) = P(- 0. 5/[pic] lt; Z lt; 0. /[pic]) = P( -0. 5/0. 2083 lt; Z lt; 0. 5/0. 2083 ) = P( -2. 4 lt; Z lt; 2. 4) = 0. 9836 b)How large a sample must be taken in order to be 98 percent certain that the sample mean is within one-half pound of the population mean? Ans: n = ? ,[pic] = /n = 1. 25/(n P(( Ã¢â‚¬â€œ 0. 5 lt; [pic] lt; ( + 0. 5) = 0. 98 ( P(-0. 5 lt; [pic]- ( lt; +0. 5) = 0. 98 ( P( -0. 5/1. 25/(n lt; Z lt; +0. 5/1. 25/(n ) = 0. 98 From Normal Table, we have P(-2. 33 lt; Z lt; 2. 33) = 0. 98 ( 0. 5/1. 25/(n = 2. 33 ( n = 33. 93 ( at least 34 households. How to cite Normal Distribution, Papers

## Saturday, April 25, 2020

### Pornoviolence By Tom Wolfe Essays - Tom Wolfe, Wolfe, Slasher Film

Pornoviolence By Tom Wolfe Judging by your most unfavorable reactions to these conversations, I can tell that you don?t like what you hear. Well, what you just heard were the first six paragraphs, the introduction of Tom Wolfe?s 1976 essay entitled Pornoviolence. This story that talks about a convention of tabloid authors begins with those banal introductions. Wolfe felt as if names were obsolete, as the authors that submit these stories are ?stringers? or correspondents from around the country and as a result, they are all known by their stories? titles. Wolfe provides a definition of pornoviolence by distinguishing it from the old pornography. He argues that the media previously attracted readership of magazines and tabloids with pornography. Instead, he claims that today pornoviolence is stimulated by violence that puts television viewers in the position of control, and he supports his position using examples of programs that contain the pornography of violence. Basically, he says that pornoviolence is the pornography of violence; only the weirdest, grossest stuff would be considered as usable material. Video games like the Mortal Kombat series, Perfect Dark, or James Bond 007 and ?slasher flicks? such as Scream, I Know What You Did Last Summer, and The Texas Chainsaw Massacre are excellent examples of gratuitous violence that have been released in the past several years. Inadvertently, Wolfe poses a question to all of us: Do we as people evolve along with the TV violence, or are television producers just giving us what they think we want? English Essays