Transcript Statistics - Nayland Maths

Statistics
Please choose a question to attempt from the following:
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STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
10
7
7
15
9
(a) Find the mean and standard deviation for this data.
(b) A similar experiment was conducted with a second
company. The results for this were….
mean = 12
and standard deviation = 1.88
How does the second company compare to the first?
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STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
(a)

SD  (b)



EXIT
10
15
9
Draw a7table 7
n
data deviation for this data.
2 comparing
Find the mean
and
standard
mean 
x  x  to mean.
 Then square
A similar experiment
was conducted with a second
n
Thevalues.
n 1
company.
results for this were….

When comparing

data sets= always
mean = 12 and standard deviation
1.88
make comment on
How does the second companythe
compare
to(which
the first?
average
is
bigger etc.) and the
spread of the data.



Reveal answer only
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STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
10
7
7
15
9
(a) Find the mean and standard deviation for this data.
(b) A similar experiment was conducted with a second
company. The results for this were….
mean = 12
and standard deviation = 1.88
How does the second company compare to the first?
(a) Mean = 8
SD  3.87
(b)The second company has a longer average response time.
The smaller standard deviation means their arrival time is
more predictable.
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Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
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1. Calculate mean.
(a)
Mean = (3+5+10+7+7+15+9)7 = 8
So x = 8
and no.pieces of data = n = 7
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
x 8
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Continue Solution
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2. Draw table comparing data to mean.
x

3
5
10
7
7
15
9
-5
-3
2
-1
-1
7
1
25
9
4
1
1
49
1
 (x - x )2 =
90
xx


xx

2
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
Begin Solution
Continue Solution
Comments
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3. Use formula to calculate standard
deviation.


xx


SD 
 n 1


 90 
SD   
 6 
SD  3.87

2





Just found!!
Question 1
(b) A similar experiment was
conducted with a second
company. The results for this
were….
mean = 12 and
standard deviation = 1.88
How does the second company
compare to the first?
x 8
1. Always compare mean and standard
deviation.
(b)The second company has a
longer average response time.
The smaller standard deviation
means their arrival time is more
predictable.
SD  3.87
What would you like to do now?
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Comments
3. Use formula to calculate standard
deviation.


xx


SD 
 n 1


 90 
SD   
 6 
SD  3.87

2





Check the list of Formulae for the
Standard Deviation Formula:
 ( x  x)
n 1
2
or
2
(
x
)
 x 2  n
n 1
The second formula can be used
in the calculator paper.
The calculator must be in
Stats. Mode to allow the data
to be entered.
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STATISTICS : Question 2
Weeks of training
A TV personality takes part in a 20 week training schedule to copy a 100m
sprinter. Her times are recorded every 2 weeks and plotted in the graph
then a line of best fit is drawn.
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STATISTICS : Question 2
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Weeks of training
(i) Find the equation of the line in terms of T and W.
(ii) Use answer (i) to predict
(a) her time after 12 weeks of training
(b) the week when her time was 11.5secs
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STATISTICS : Question 2
Find
gradient
Use
Use
your
your and
note
intercept
equation
equation
and
and of
T axis.
substitute
substitute
WT= =12.
11.5
Reveal answer only
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Weeks of training
(i) Find the equation of the line in terms of T and W.
(ii) Use answer (i) to predict
(a) her time after 12 weeks of training
(b) the week when her time was 11.5secs
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STATISTICS : Question 2
What would you like to do now?
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Weeks of training
(i) Find the equation of the line in terms of T and W.
(ii) Use answer (i) to predict
EXIT
T = -¼ W + 15
(a) her time after 12 weeks of training
12secs.
(b) the week when her time was 11.5secs
14 weeks
Question 2
(i) Find the equation of the line
in terms of T and W.
1. Find gradient and note intercept of
T axis.
(a) m = (y2 – y1) 10 - 15
= -¼
=
(x2 – x1)
20 - 0
Intercept at 15
Equation is
Go to graph
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T = -¼ W + 15
Question 2
(ii) Use answer (i) to predict
(a) her time after 12 weeks of
training
(b) the week when her time was
11.5secs
2. Use your equation and substitute
W = 12.
(b)(i) If w = 12 then T = -¼ W + 15
becomes
T = (-¼ x 12) + 15
= -3 + 15
= 12
Time at 12 weeks is 12secs.
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Question 2
(ii) Use answer (i) to predict
(a) her time after 12 weeks of
training
(b) the week when her time was
11.5secs
3. Use your equation and substitute
T = 11.5
(b)(i) If t = 11.5 then
becomes
T = -¼ W + 15
-¼ W + 15 = 11.5
(-15) (-15)
-¼ W = -3.5 x (–4)
W = 14
Reach a time of 11.5sec after 14 weeks.
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Comments
1. Find gradient and note intercept
of T axis.
(a)
m = (y2 – y1) 10 - 15
= -¼
=
(x2 – x1)
20 - 0
To find the equation of a
line from the graph:
Must Learn:
y = mx + c
Intercept at 15
Equation is
T = -¼ W + 15
gradient
intercept
So you need to find these!!
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Comments
1. Find gradient and note intercept
of T axis.
(a)
m = (y2 – y1) 10 - 15
= -¼
=
(x2 – x1)
20 - 0
vertical
m =
horizontal
Note: Always draw the
horizontal before the vertical:
Intercept at 15
horizontal (+ve)
Equation is
T = -¼ W + 15
vertical (-ve)
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STATISTICS : Question 3
A sample of 180 teenagers were asked their opinions on the TV series the
“Simpsons” & the movie “Shrek” and their responses were displayed in
the following table…
Like
Don’t like
Simpsons
Like
Shrek
Don’t like
Shrek
Simpsons
126
?
15
21
(a) What percentage liked Shrek but not the Simpsons?
(b) If someone is picked at random what is the probability that
(i) they liked the Simpsons but not Shrek?
(ii) they liked neither?
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Statistics Menu
STATISTICS : Question 3
A sample of 180 teenagers were asked their opinions on the TV series the
“Simpsons” & the movie “Shrek” and their responses were displayed in
the following table…
Like
Don’t like
Simpsons
Like
Shrek
Don’t like
Shrek
126
15
Simpsons
Remember
To find probabilities
all entries in
use:
? P = no oftable must
add to total
favourable / no of
sample size
data
21
(a) What percentage liked Shrek but not the Simpsons?
(b) If someone is picked at random what is the probability that
(i) they liked the Simpsons but not Shrek?
(ii) they liked neither?
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Full solution
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Reveal
STATISTICS : Question 3
A sample of 180 teenagers were asked their opinions on the TV series the
“Simpsons” & the movie “Shrek” and their responses were displayed in
the following table…
Like
Don’t like
Simpsons
Like
Shrek
Don’t like
Shrek
Simpsons
126
?
15
21
(a) What percentage liked Shrek but not the Simpsons?
10%
(b) If someone is picked at random what is the probability that
(i) they liked the Simpsons but not Shrek?
(ii) they liked neither?
Full solution
EXIT
Comments
7/
60
Statistics Menu
What now?
1/
12
Question 3
1. Use P = no of favourable / no of data
Like
Simpsons
Like
Shrek
Don’t like
Shrek
Don’t like
Simpsons
126
?
15
21
(a) What percentage liked Shrek
but not the Simpsons?
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NB: There are 180 in survey!!
(a)
Like Shrek but not Simpsons
= 180 – 126 – 15 – 21
= 18
% = 18/180 =
1/
10
= 10%
Question 3
1. Use P = no of favourable / no of data
Like
Simpsons
Like
Shrek
Don’t like
Shrek
Don’t like
Simpsons
126
?
15
21
what is the probability that
(i) they liked the Simpsons but
not Shrek?
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(b)(i) Prob =
15/
180
=
1/
12
Question 3
1. Use P = no of favourable / no of data
Like
Simpsons
Like
Shrek
Don’t like
Shrek
Don’t like
Simpsons
126
?
15
21
(b)(ii) Prob =
21/
180
=
7/
60
What would you like to do now?
(ii) they liked neither?
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Comments
1. Use P = no of favourable / no of data
NB: There are 180 in survey!!
(a)
Like Shrek but not Simpsons
Note: To change a fraction to
a % multiply by 100%
18
180
=
18
180
x 100%
= 180 – 126 – 15 – 21
= 18
% = 18/180 =
1/
10
=
10%
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Comments
To calculate simple probabilities:
1. Use P = no of favourable / no of data
Probability =
(b)(i) Prob =
15/
180
=
1/
12
Number of favourable outcomes
Number of possible outcomes
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STATISTICS : Question 4
Get hint
On a college course you have to pick a language plus a
Reveal ans
leisure activity from the following lists
LANGUAGE
FRENCH
GERMAN
SPANISH
LEISURE
MUSIC
VIDEO
PRODUCTION
BASKETBALL
SWIMMING
Full solution
Comments
Stats Menu
(a) Make a list of all possible combinations of courses.
(b) If a combination is selected at random what is the
probability that it is ……
(i)
Includes Spanish?
(ii) Includes swimming?
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(iii) Doesn’t include French but includes music?
What would you like to do now?
STATISTICS : Question 4
On a college course you have to pick a language plus a
Reveal ans
leisure activity from the following lists
LANGUAGE
FRENCH
GERMAN
SPANISH
LEISURE
MUSIC
VIDEO
PRODUCTION
BASKETBALL
SWIMMING
Full solution
Comments
Stats Menu
(a) Make a list of all possible combinations of courses. Use a tree
diagram &
“branch
(b) If a combination is selected at random what is the Now
list
out”
with
probability that it is ……
the pairs
each
Use your
From list
eachof
of
language.
language
“branch
(i) Includes Spanish?
possible
subjects.
out” with each
combinations
to find
(ii) Includes swimming?
leisure activity.
probabilities.
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(iii) Doesn’t include French but includes music?
STATISTICS : Question 4
On a college course you have to pick a language plus a
leisure activity from the following lists
LANGUAGE
FRENCH
GERMAN
SPANISH
Full solution
LEISURE
MUSIC
VIDEO
PRODUCTION
BASKETBALL
SWIMMING
(a) Make a list of all possible combinations of courses.
Comments
Stats Menu
CLICK
(b) If a combination is selected at random what is the
probability that it is ……
(i)
Includes Spanish?
(ii) Includes swimming?
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1/
3
1/
4
1/
(iii) Doesn’t include French but includes music?
6
(a)
French/Music
French/Video
French/Bsktbll
French/Swim
German/Music
GERMAN
German/Video
German/Bsktbll
German/Swim
Spanish/Music
Spanish/Video
Spanish/Bsktbll
Spanish/Swim
Hints
Use a tree diagram &
“branch out” with
each language.
From each language
“branch out” with
each leisure activity.
Now list the pairs of
subjects.
Hints
French/Music
Have list of combinations
handy.
French/Video
French/Bsktbll
(b)(i) Prob = 4/12 =
1/
3
Remember to simplify.
French/Swim
What now?
German/Music
German/Video
Comments
(b)(ii) Prob = 3/12 =
1/
4
Stats Menu
German/Bsktbll
What is probability:
German/Swim
(i)
Includes Spanish?
Spanish/Music
Spanish/Video
Spanish/Bsktbll
Spanish/Swim
(ii) Includes swimming?
(b)(iii) Prob = 2/12 = 1/6
(iii)Doesn’t include French
but includes music?
Comments
(a)
French/Music
French/Video
No. of possible
outcomes
French/Bsktbll
French/Swim
GERMAN
= 3 x 4 = 12
German/Music
To calculate simple
probabilities:
German/Video
Probability =
German/Bsktbll
German/Swim
Number of favourable
Number of possible
Spanish/Music
Spanish/Video
Spanish/Bsktbll
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Spanish/Swim
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STATISTICS : Question 5
The delivery times for a fast food company are shown in the
following cumulative frequency table.
Time
No. Deliveries
up to 4 mins
up to 8 mins
up to 12 mins
5
13
37
up to 16 mins
up to 20 mins
53
60
Get hint
Reveal ans
Full solution
Comments
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(a)
How many deliveries took longer than 16 mins?
(b)
Use the data to construct a cumulative frequency graph.
(c)
Use the graph to find the median and semi-interquartile
range for this data.
STATISTICS : Question 5
The delivery times for a fast food company are shown in the
following cumulative frequency table.
What would you like to do now?
For how many
Time
greater than
Median is
16 up
findto 4 mins
middle
difference
valueso 8we
up to
mins
between
end
SIQR
=
½(Q3
– Q1)
want
value
of
up to 12 mins
halfwayvalues
to point &
Q1up
– 25%
point
on
up–to
16point
mins
16.
Q3
75%
frequency
up
to 20 mins
axis.
No. Deliveries
5
13
37
53
60
Reveal ans
Full solution
Comments
Stats Menu
(a)
How many deliveries took longer than 16 mins?
(b)
Use the data to construct a cumulative frequency graph.
(c)
Use the graph to find the median and semi-interquartile
range for this data.
STATISTICS : Question 5
The delivery times for a fast food company are shown in the
following cumulative frequency table.
Time
No. Deliveries
up to 4 mins
up to 8 mins
up to 12 mins
5
13
37
up to 16 mins
up to 20 mins
53
60
Full solution
Comments
Stats Menu
=7
(a)
How many deliveries took longer than 16 mins?
(b)
Use the data to construct a cumulative frequency graph.
(c)
Use the graph to find the median and semi-interquartile
range for this data. Median = 11 mins SIQR = 2.75 mins
(a) Deliveries taking longer than 16 mins = 60 – 53 = 7
Time Deliveries
Cum freq
4
5
¾ of 60 = 45
Q3
Q2
8
13
12
37
Comments
½ of 60 = 30
16
53
20
60
Stats Menu
¼ of 60 = 15
What would you like to do now?
Q1
8.5
(C) Median = 11mins.
11
14
Delivery time(mins)
(c) SIQR = ½(Q3 – Q1) = (14 – 8.5)  2 = 2.75mins
Comments
Note:
In a Cumulative Frequency
Diagram:
Cumulative
Frequency
On y-axis (cumulative
frequency)
60
Q1 at 25%,
50%
Q2 at 50%,
Q3 at 75%
11
20
Delivery
Time
And read values from the
delivery time scale (x-axis).
End of Statistics
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