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Date May 2018 Marks available 3 Reference code 18M.2.SL.TZ1.S_8
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Estimate Question number S_8 Adapted from N/A

Question

The following table shows values of ln x and ln y.

The relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.

Find the value of a and of b.

[3]
a.

Use the regression equation to estimate the value of y when x = 3.57.

[3]
b.

The relationship between x and y can be modelled using the formula y = kxn, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.

[7]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach      (M1)

eg  one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14      A1A1 N3

[3 marks]

a.

correct substitution     (A1)

eg   −0.454 ln 3.57 + 6.14

correct working     (A1)

eg  ln y = 5.56484

261.083 (260.409 from 3 sf)

y = 261, (y = 260 from 3sf)       A1 N3

Note: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.

[3 marks]

b.

METHOD 1

valid approach for expressing ln y in terms of ln x      (M1)

eg  ln y = ln ( k x n ) , ln ( k x n ) = a ln x + b

correct application of addition rule for logs      (A1)

eg  ln k + ln ( x n )

correct application of exponent rule for logs       A1

eg  ln k + n ln x

comparing one term with regression equation (check FT)      (M1)

eg   n = a , b = ln k

correct working for k      (A1)

eg   ln k = 6.14210 , k = e 6.14210

465.030

n = 0.454 , k = 465  (464 from 3sf)     A1A1 N2N2

 

METHOD 2

valid approach      (M1)

eg   e ln y = e a ln x + b

correct use of exponent laws for  e a ln x + b      (A1)

eg   e a ln x × e b

correct application of exponent rule for  a ln x      (A1)

eg   ln x a

correct equation in y      A1

eg   y = x a × e b

comparing one term with equation of model (check FT)      (M1)

eg   k = e b , n = a

465.030

n = 0.454 , k = 465 (464 from 3sf)     A1A1 N2N2

 

METHOD 3

valid approach for expressing ln y in terms of ln x (seen anywhere)      (M1)

eg   ln y = ln ( k x n ) , ln ( k x n ) = a ln x + b

correct application of exponent rule for logs (seen anywhere)      (A1)

eg   ln ( x a ) + b

correct working for b (seen anywhere)      (A1)

eg   b = ln ( e b )

correct application of addition rule for logs      A1

eg   ln ( e b x a )

comparing one term with equation of model (check FT)     (M1)

eg   k = e b , n = a

465.030

n = 0.454 , k = 465 (464 from 3sf)     A1A1 N2N2

[7 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 4—Statistics and probability » SL 4.4—Pearsons, scatter diagrams, eqn of y on x
Show 146 related questions
Topic 1—Number and algebra » SL 1.5—Intro to logs
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Topic 1—Number and algebra
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