Find the following probabilities by checking the z table i) P((Z>-0.8) ii) P (-1.3iii) Z0.2 Part b) A new car has recently hit the market. The distance travelled on 1 gallon of fuel is normally...

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Find the following probabilities by checking the z table
i) P((Z>-0.8)
ii) P (-1.3iii) Z0.2
Part b)
A new car has recently hit the market. The distance travelled on 1 gallon of fuel is
normally distributed with a mean of 65 miles and a standard deviation of 4 miles. Find
the probability of the following events.
i) The car travels more than 70 miles per gallon.
ii) The car travels less than 60 miles per gallon.
iii) The car travels between 55 and 70 miles per gallon.
Question 2 [3 marks]
Part a)
A sample of n=25 observations is drawn from a normal population with µ=100 and
s=20. Find the following.
i) P( ii) P(96Part b)
The amount of time the university professors devote to their jobs per week is normally
distributed with a mean of 52 hours and a standard deviation of 6 hours.
i) What is the probability that a professor works for more than 60 hours per weeks?
ii) Find the probability that the mean amount of work per week for three randomly


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Individual Assignment Unit: TSTA101 – INTRODUCTORY STATISTICS Due Date: Thursday, 28/09/2017 (16:00pm) Number of Questions: Six (6) Questions Total Marks: Twenty (20) marks Instructions: All questions should be attempted. The marks of each question would be awarded based on your understanding of the questions, concepts and procedures; hence you should demonstrate your answers step by step.Question 1 [3 marks] Part a) Find the following probabilities by checking the z table i) P((Z>-0.8) ii) P (-1.3<-0.7)><96)><105)>






Individual Assignment Unit: TSTA101 – INTRODUCTORY STATISTICS Due Date: Thursday, 28/09/2017 (16:00pm) Number of Questions: Six (6) Questions Total Marks: Twenty (20) marks Instructions: All questions should be attempted. The marks of each question would be awarded based on your understanding of the questions, concepts and procedures; hence you should demonstrate your answers step by step. Question 1 [3 marks] Part a) Find the following probabilities by checking the z table i) P((Z>-0.8) ii) P (-1.3<><-0.7) iii)="" z0.2="" part="" b)="" a="" new="" car="" has="" recently="" hit="" the="" market.="" the="" distance="" travelled="" on="" 1="" gallon="" of="" fuel="" is="" normally="" distributed="" with="" a="" mean="" of="" 65="" miles="" and="" a="" standard="" deviation="" of="" 4="" miles.="" find="" the="" probability="" of="" the="" following="" events.="" i)="" the="" car="" travels="" more="" than="" 70="" miles="" per="" gallon.="" ii)="" the="" car="" travels="" less="" than="" 60="" miles="" per="" gallon.="" iii)="" the="" car="" travels="" between="" 55="" and="" 70="" miles="" per="" gallon.="" question="" 2="" [3="" marks]="" part="" a)="" a="" sample="" of="" n="25" observations="" is="" drawn="" from="" a="" normal="" population="" with="" µ="100" and="" σ="20." find="" the="" following.="" i)="" p(=""><96) ii)=""><><105) part b) the amount of time the university professors devote to their jobs per week is normally distributed with a mean of 52 hours and a standard deviation of 6 hours. i) what is the probability that a professor works for more than 60 hours per weeks? ii) find the probability that the mean amount of work per week for three randomly selected professors is more than 60 hours? question 3 [2 marks] part a) given the following information =500, σ=12, n=50 i) determine the 95% confidence interval estimate of population mean. ii) determine the 99% confidence interval estimate of population mean. part b) a statistics practitioner calculated the mean and standard deviation from a sample of 51. they are =120 and s=15. (i) estimate the population mean with 95% confidence level. (ii) estimate the population mean with 99% confidence level. x x x x question 4 [4 marks] part a) calculate the statistic, set up the rejection region, interpret the result, and draw the sampling distribution. h0: µ=10 h1: µ≠10 given that: σ=10, n=100, =10, α=0.05. part b) a statistics practitioner is in the process of testing to determine whether is enough evidence to infer that the population mean is different from 180. she calculated the mean and standard deviation of a sample of 200 observations as =175 and s=22. calculate the value of the test statistic of the test required to determine whether there is enough evidence to infer at the 5% significance level that the population mean is different from 180. question 5 (2 marks) suppose you are using a completely randomized design to study some phenomenon. there are five treatment levels and a total of 55 people in the study. each treatment level has the same sample size. complete the following anova. use α=0.05 to find the table f value and use the data to test the null hypothesis. source of variance ss df ms f treatment 583.39 error 972.18 total 1555.57 question 6 (6marks) there is a simple linear regression model given by: where price = used car price in dollars and age = age of the car in years. the excel results obtained using ordinary least squares are presented below: regression statistics r2 0.077 standard error 42069 observations 117 coefficients standard error t stat intercept a 6748 7.035 age -2658 856 b use the above output for answering the following questions: a) calculate the missing values from the summary output: a and b b) interpret the slope of the regression line. c) write down the estimated linear regression line. d) what is the value of the coefficient of determination? interpret this value e) what is the value of the coefficient of correlation? interpret this value. f) test whether the estimated coefficient of age is significantly less than zero at the 5% level of significance. g) predict price if the car has driven 3 years. x x part="" b)="" the="" amount="" of="" time="" the="" university="" professors="" devote="" to="" their="" jobs="" per="" week="" is="" normally="" distributed="" with="" a="" mean="" of="" 52="" hours="" and="" a="" standard="" deviation="" of="" 6="" hours.="" i)="" what="" is="" the="" probability="" that="" a="" professor="" works="" for="" more="" than="" 60="" hours="" per="" weeks?="" ii)="" find="" the="" probability="" that="" the="" mean="" amount="" of="" work="" per="" week="" for="" three="" randomly="" selected="" professors="" is="" more="" than="" 60="" hours?="" question="" 3="" [2="" marks]="" part="" a)="" given="" the="" following="" information="500," σ="12," n="50" i)="" determine="" the="" 95%="" confidence="" interval="" estimate="" of="" population="" mean.="" ii)="" determine="" the="" 99%="" confidence="" interval="" estimate="" of="" population="" mean.="" part="" b)="" a="" statistics="" practitioner="" calculated="" the="" mean="" and="" standard="" deviation="" from="" a="" sample="" of="" 51.="" they="" are="120" and="" s="15." (i)="" estimate="" the="" population="" mean="" with="" 95%="" confidence="" level.="" (ii)="" estimate="" the="" population="" mean="" with="" 99%="" confidence="" level.="" x="" x="" x="" x="" question="" 4="" [4="" marks]="" part="" a)="" calculate="" the="" statistic,="" set="" up="" the="" rejection="" region,="" interpret="" the="" result,="" and="" draw="" the="" sampling="" distribution.="" h0:="" µ="10" h1:="" µ≠10="" given="" that:="" σ="10," n="100," =10,="" α="0.05." part="" b)="" a="" statistics="" practitioner="" is="" in="" the="" process="" of="" testing="" to="" determine="" whether="" is="" enough="" evidence="" to="" infer="" that="" the="" population="" mean="" is="" different="" from="" 180.="" she="" calculated="" the="" mean="" and="" standard="" deviation="" of="" a="" sample="" of="" 200="" observations="" as="175" and="" s="22." calculate="" the="" value="" of="" the="" test="" statistic="" of="" the="" test="" required="" to="" determine="" whether="" there="" is="" enough="" evidence="" to="" infer="" at="" the="" 5%="" significance="" level="" that="" the="" population="" mean="" is="" different="" from="" 180.="" question="" 5="" (2="" marks)="" suppose="" you="" are="" using="" a="" completely="" randomized="" design="" to="" study="" some="" phenomenon.="" there="" are="" five="" treatment="" levels="" and="" a="" total="" of="" 55="" people="" in="" the="" study.="" each="" treatment="" level="" has="" the="" same="" sample="" size.="" complete="" the="" following="" anova.="" use="" α="0.05" to="" find="" the="" table="" f="" value="" and="" use="" the="" data="" to="" test="" the="" null="" hypothesis.="" source="" of="" variance="" ss="" df="" ms="" f="" treatment="" 583.39="" error="" 972.18="" total="" 1555.57="" question="" 6="" (6marks)="" there="" is="" a="" simple="" linear="" regression="" model="" given="" by:="" where="" price="used" car="" price="" in="" dollars="" and="" age="age" of="" the="" car="" in="" years.="" the="" excel="" results="" obtained="" using="" ordinary="" least="" squares="" are="" presented="" below:="" regression="" statistics="" r2="" 0.077="" standard="" error="" 42069="" observations="" 117="" coefficients="" standard="" error="" t="" stat="" intercept="" a="" 6748="" 7.035="" age="" -2658="" 856="" b="" use="" the="" above="" output="" for="" answering="" the="" following="" questions:="" a)="" calculate="" the="" missing="" values="" from="" the="" summary="" output:="" a="" and="" b="" b)="" interpret="" the="" slope="" of="" the="" regression="" line.="" c)="" write="" down="" the="" estimated="" linear="" regression="" line.="" d)="" what="" is="" the="" value="" of="" the="" coefficient="" of="" determination?="" interpret="" this="" value="" e)="" what="" is="" the="" value="" of="" the="" coefficient="" of="" correlation?="" interpret="" this="" value.="" f)="" test="" whether="" the="" estimated="" coefficient="" of="" age="" is="" significantly="" less="" than="" zero="" at="" the="" 5%="" level="" of="" significance.="" g)="" predict="" price="" if="" the="" car="" has="" driven="" 3="" years.="" x="">
Answered Same DayDec 27, 2021

Answer To: Find the following probabilities by checking the z table i) P((Z>-0.8) ii) P (-1.3iii) Z0.2 Part b)...

David answered on Dec 27 2021
139 Votes
Question 1
Part A
I Find: P (Z > −0.8)
Area to the right of −0.80
Z
0.7881
−0.80
0
P (Z > −0.8) = 1− P (Z < −0.8)
P (Z < −0.8): in a z-table having area to the left of z, locate -0.8 in the left most
column. Move across the row to the right under column 0.00 and get value 0.2119
P (Z > −0.8) = 1− P (Z < −0.8) = 1− 0.2119 = 0.7881
P (Z > −0.8) = 0.7881
Using excel function 1-normdist(0.8, 0, 1, TRUE) or TI’s function
normalcdf(-0.8,1E99 ) answer is: 0.788144601416603
II Find: P (
−1.3 < Z < −0.7)
Area in between −1.30 and −0.70
Z
0.0968
−0.70
−1.30
0
0.2420
0.2420− 0.0968 = 0.1452
P (−1.3 < Z < −0.7) = P (Z < −0.7)− P (Z < −1.3)
P (Z < −0.7): in a z-table having area to the left of z, locate -0.7 in the left most
column. Move across the row to the right under column 0.00 and get value 0.2420
P (Z < −1.3): in a z-table having area to the left of z, locate -1.3 in the left most
column. Move across the row to the right under column 0.00 and get value 0.0968
P (−1.3 < Z < −0.7) = 0.2420− 0.0968 = 0.1452
P (−1.3 < Z < −0.7) = 0.1452
Note that exact answer using technology is:
0.145163167637463
III Find: z0.2
P (Z > zα) = 0.2 =⇒ P (Z < zα) = 0.8
Z
0.8
z0.2
0
(0.8000) is not present in the z-table containing area to left of z, but we can use
linear interpolation.
We note that nearby 0.7995 has z-score 0.84 and 0.8023 has z-score 0.85.
Thus we can use “linear interpolation” to find out approximate z-score = 0.842
0.8000 is 0.8000− 0.80230.7995− 0.8023 = 23/28 way from 0.8023 to 0.7995.
So, the z-score for 0.8000 is approximately 23/28 of the way from 0.85 to 0.84
Hence:
z-score = 0.85 + ((0.84)− (0.85)) ∗ 23/28 = 0.8418
Bonus fact: Using TI-83/84 function invNorm(0.8) or Excel function: NormInv(0.8, 0, 1),
exact z-score of 0.8 is 0.8416212336
Thus z0.2 = 0.842
Critical value: z0.2 = 0.842
Part B
I P (X > 70)
Normal Distribution, µ = 65, σ = 4, we convert this to standard normal using
z = x− µ
σ
z = 70− (65)4 = 1.25
P (Z > 1.25) = Area to the right of 1.25
1
Z
0.1056
1.25
0
P (X > 70) = P (Z > 1.25) = 1− P (Z < 1.25) = 1− 0.8944
= 0.1056
P (Z < 1.25): in a z-table having area to the left of z, locate 1.2 in the left most column. Move
across the row to the right under column 0.05 and get value 0.8944
Using technology, answer is: 0.105649773666855
P (X > 70) = 0.1056 = 10.56%
II P (X < 60)
Normal Distribution, µ = 65, σ = 4, we convert this to standard normal using
z = x− µ
σ
z = 60− (65)4 = −1.25
P (Z < −1.25) = Area to the left of −1.25
Z
0.1056
−1.25
0
P (X < 60) = P (Z < −1.25) = 0.1056 (from z-table)
P (Z < −1.25): in a z-table having area to the left of z, locate -1.2 in the left most column. Move
across the row to the right under column 0.05 and get value 0.1056
Using technology, answer is: 0.105649773666855
P (X < 60) = 0.1056 = 10.56%
III P (55 < X < 70)
Normal Distribution, µ = 65, σ = 4, we convert this to standard normal using
z = x− µ
σ
z1 =
55− (65)
4 = −2.50
z2 =
70− (65)
4 = 1.25
P (−2.50 < Z < 1.25) = Area in between −2.50 and 1.25
Z
0.0062
1.25
−2.50
0
0.8944
0.8944− 0.0062 = 0.8882
P (55 < X < 70) = P (−2.50 < Z < 1.25)
= P (Z < 1.25)− P (Z < −2.50) = 0.8944− 0.0062= 0.8882
P (Z < 1.25): in a z-table having area to the left of z, locate 1.2 in the left most column. Move
across the row to the right under column 0.05 and get value 0.8944
P (Z < −2.5): in a z-table having area to the left of z, locate -2.5 in the left most column. Move
across the row to the right under column 0.00 and get value 0.0062
Using technology, answer: 0.888140561007369
P (55 < X < 70) = 0.8882 = 88.82%
\ZXY[
Question 2
Part A
I P (X < 96)
Normal Distribution, µ = 100, σ = 20, n = 25
we convert this to standard normal using z = x− µx
σx
= x− µ
σ/

n
z = 96− (100)
20/

25
= −1.00
P (Z < −1.00) = Area to the left of −1.00
2
Z
0.1587
−1.00
0
P (X < 96) = P (Z < −1.00) = 0.1587 (from z-table)
P (Z < −1): in a z-table having area to the left of z, locate -1 in the left most column. Move
across the row to the right under column 0.00 and get value 0.1587
Using technology, answer is: 0.158655253931457
P (X < 96) = 0.1587 = 15.87%
II P (96 Normal Distribution, µ = 100, σ = 20, n = 25
we convert this to standard normal using z = x− µx
σx
= x− µ
σ/

n
z1 =
96− (100)
20/

25
= −1.00
z2 =
105− (100)
20/

25
= 1.25
P (−1.00 < Z < 1.25) = Area in between −1.00 and 1.25
Z
0.1587
1.25
−1.00
0
0.8944
0.8944− 0.1587 = 0.7357
P (96 = P (Z < 1.25)− P (Z < −1.00) = 0.8944− 0.1587= 0.7357
P (Z < 1.25): in a z-table having area to the left of z, locate 1.2 in the left most column. Move
across the row to the right under column 0.05 and get value 0.8944
P (Z < −1): in a z-table having area to the left of z, locate -1 in the left most column. Move
across the row to the right under column 0.00 and get value 0.1587
Using...
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