Need help solving Probability and Statistics questions

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InEk7

Mathematics

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Please help me solve the questions and provide me solutions on how to solve them. I have attached the queestions in the word doc.

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Explanation & Answer

know Can you please confirm if you have received the work? Once again, thanks for allowing me to help you R MESSAGE TO STUDYPOOL NO OUTLINE IS NEEDED AS IT IS A MULTIPLE CHOICES I AM 100% confirm that all question are right

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60 - 70
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80 - 90
90 -100

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Frequency
Frequency

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Answer :21%
Working :Percentage of visitors who brought sack lunch= (no. of visitors who brought sack lunch/ total no. of visitors) *100
= (123/(123+256+208) * 100= (123/587) *100
= 20.95400341%
=21% ( rounded to the nearest interger)
=

Answer: The first graph dispalys the number of students favoring each color while the second displays
the percentages for each color.

answer ;the second choice; the graph is symmetrical , having frequencies; 1, 4, 5,4 and 1 for
the values 1, 2 , 3 , 4 and 5.

Answer :The range for group A is smaller than the range for group B.

Explanation; range is the difference between the maximum value and the minimum value of a data
.From the side by side box plots , it clear that the minimum for group A is close the median of group B
and that both groups have the same maximum value . thus group A has a smaller range than group B.
Box plots donot tells us about outliers, the number of individuals , but show the distribution across the
group .Also , not all shoppers in group A spent more time in the store than 50% of the shoppers from
agroup B since the minimum value in group A is even less than 50% of group B. So, the second , third
and fourth choice are not correct.

Answer: there are no outliers

ANSWER:The distribution is skewed to the left .

Answer. The mean and not the median, because the mean is 10 while the median is low at 6

Answer : the mean of B is higher than the mean of A: B is left- skewed means that high number of data
values are concentrated to upper side of distribution and its mean is high.Unlike left-skewed , right –
skewed high number of data are concentrated on the lower side of the distribution and hence the sum
total of these values is low . Therefore , its mean will be lower than that of B.

Answer: The three shifts have the same interquartile range
Working; interquartile range = Q3-Q1 ; Shift A; Q3-Q1= 26-19= 7, Shift B; Q3-Q1= 25-18= 7,
Shift C; Q3-Q1= 27-20=7

Answer :the median level of vitamin c in the bloodstream is higher for those who drank orange juice
Explanation:From the boxplots , the median level of vitamin for those drank orange juice is between
20% - 25% while those who took capsules is about slightly above 15%. Thus , the median of median level
of vitamin for those drank orange juice is higher than those took capsules. Also , vitamin c taken from
orange juice tends to stay in the bloodstream longer than vitamin c from capsules.In addition , the amount of
vitamin c in the blood stream is not consistent for those who took capsules as it can be seen in the blox plot.

Answer ;Between 6.05 and 7.55 hours

Workings
68% Confidence interval
Z score for 68% = 1 or 68% is 1 standard deviation frocm the mean
68% Confidence interval
=(6.8- 1*0.75, 6.8+1*0.75)
=(6.05,7.55)

Answer: 2nd choice ; the high school classroom is the explanatory variable (x) and the type of instruction
is the response variable .

Answer : the 1st choice : it appears that students who took math as seniors in high school graduated
college with higher GPAs than those who did not take math , but there is also greater spread in the
GPA’s of college students who did not take math as high school seniors .
Explanation :The spread of the data can be depicted from the range , maximum – minimum . The range
for students who took math is 1.7 , ( 4.0-2.3=1.7) , while the range for those who did not take math is
2,(3.9-1.9=2).All the values for the minimum , Q1, median , Q3 , maximum for the 1000students who
took math are higher than for those who didn’t take math.

Answer;third graph or third choice

Answer:the 2nd choice or 2nd table ; percentages are based on each gender , separately .

Answer: Positive linear relationship with outlier(s).
Explanation .the data points appear to be in a line, except for one point that lies outside the range of the
data points, the outlier. As temperatures of the days increases the number of ice cream cans sold
increase hence the two variables are directly proportional .Thus a positive linear relationship exist
between the number of ice cream and temperatures number of ice cream of the day.

Answer: If the outlier is removed , r would increase.

Answers: There is a strong relationship between the wait time and the eruption for the old Faithful
geyser . Also, an eruption duration of 2.4 minutes predicts a wait time of 60.0 ,minutes
Explanation & working
The correlation coefficient is both positive and close to 1 , which implies a strong linear relationships
between the two variables , the wait time and the eruption of geyser .
By plugging values in the regression equation ;y=11.80x+31.68

X=1min;y=11.80*1+31.68=43.48 mins
X=2.4 mins;y=11.80*2.4+31.68= 60 mins,
x=5.0;y=11.80*5+31.68=90.68 mins,
x=3.5 mins; 11.80*3.5+31.68 = 72.98mins
So, only the 1st and 3rd choices are correct.

Answer: The 1ST CHOICE.
Explanation ; this is due to low percentages of acceptance and low number of men who were accepted.

Answers:The variable ‘’number of churches’’ can be used to only predict the crime rate .Also , the
population increase for the city over the past 50 years is a lurking variable that could influence both the
crime rate and the number of churches to increase.

Answer:Weaather conditions of the destinations is a lurking variable in this situation, since this could
t...


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