UNIVERSITY OF PRETORIA DEPARTMENT OF GEOLOGY

GEOSTATISTICS AND ORE RESERVE CALCULATIONS

GLY 362

EXAM June 2006

Examiners: Dr F. Camisani-Calzolari Time: 1.5 hrs

Dr. K T Witthüser Marks: 90

Answer all the questions. Good luck!

1. Define or explain [40]

a. Inferred and Measured Mineral Resource including statistical

measures for their classification. [10]

b. Volume variance effect [5]

c. Lognormal Kriging [5]

d. Kriging for non-stationary sample values [7]

e. Quartiles [5]

f. Sketch an Exponential semivariogram model and explain nugget

effect, range and sill. [8]

2. 20 boreholes were drilled through a gold deposit and analyzed in bulk for gold. The

distribution of the Au values in g/t was found to be lognormal. The mean of the

logarithmic (loge) transform of the concentration data of gold is 1.8 (loge g/t) and the

small sample variance, s2, of the logarithmic (loge) transform of the concentration

data of gold is 0.78947. Calculate the best estimate of the average gold concentration

in the deposit and give the 90% confidence limits. [15]

3. Evaluate P(16<x<24) and P(x>30.32) for an n(20, 16) population. [5]

4. Could a sample of a gold ore with bar-x = 15.6 ppm Au (s2 = 9 and n = 11) be

regarded as a sample of a population with µ0 = 13.5 ppm Au, at the 95% level of

confidence? [10]

5. A mineralogist did point counting on a gold ore deposit with an assay grade of 8 g/t.

How many gold grains can he expect to find after counting 2 polished sections of

solid rock, each containing 5x106 grains. What is the probability to observe only 2

grains? Assume the density of gold to be 18g/cc and that of the rock 2.7 g/cc, that all

the grains form perfect spheres with a diameter of 12 micron and that the gold grains

are randomly distributed through the ore. [10]

6. A groundwater exploration campaign has a success ratio of 30 in 100. What is the

probability of drilling one or more successful holes in a drilling campaign of 6

boreholes? [10]

Equations:

Arithmetic mean: ∑=

=n

i

ixn

x1

1

Variance: ( ) nnsxxn

V is /)1(1 222 −=−== ∑σ

Small sample variance: )1/(Vs

2 −= nns

Z-transformation: σ

µ -x =Z

Binomial distribution:

tt

t

x-nxn

xt p)-(1 p )( =)x=P(x =(x) f

∑≤t

1 = i

x-nxn

xtttt

tp)-(1 p )( =)xP(x =)F(x

)!(!

!)(n

x t

tt xnx

n

−=

E(x)= µ = nπ

Poisson distribution: pt(x=k) = [( λt)k exp(-λt)]/k! =[(µ)k exp(-µ)]/k!

E(x) = µ = λt

Exponential distribution:

f(x) = (1/θ) exp(-x / θ)

F(x)= 1-exp(-x / θ)

E(x) = µ = θ = 1/ λt

T-test statistic: es

x

ns

xT

−=

−= 00 µµ

follows t(n-1) distribution.

Confidence limits

for µ: n

stx

n

stx ⋅+<<⋅− νανα µ ,2/,2/

Selection calculations:

)(*

**z

pc ϕ

σµµ +=

( )

at x curve normal standard ofheight )(

fraction as cutoff above Proportion p

sample from estimated grade Average

cutoff above grade Average

5.0exp2

15.0exp

2

1)(

cutoff

*

*

2

*

*

_2

=

=

=

=

−−=−=

z

xzz

c

offcut

ϕ

µ

µ

σ

µ

ππϕ

Lognormal distribution

(y = log-transformed

data),

for n > 40:

Estimated mean of the raw data:

)5.0exp( 2*sye +== ττ

Estimated variance of the raw data:

( )1)exp( 22 −= sτω

Lognormal distribution

(y = log-transformed

data),

for small samples:

Estimated mean of the raw data: ( ) )exp( t sSichel'*yVsnγτ ==

Confidence intervals:

),(),( _.0_.0 nVnV supperxxslowerxx ψττψτ ⋅<<⋅

Spherical Semivariogram

model:

0)0 ,( Chh ==γ

CCahh

a

h

a

hCCahh

+=>

−⋅+=<<

0

3

3

0

) ,(

2

2

3)0 ,(

γ

γ

Ordinary Kriging:

∑=

=n

i

ii xwT1

*

( )∑=

−+=n

i

nnOK TTTxw1

),(),( γλγσ

Tab. 1: Percentage points of the t-distribution:

10 % 5 % 2.5 %

ν = df = 9 1.383 1.833 2.262

10 1.372 1.812 2.228

11 1.363 1.796 2.201

24 1.318 1.711 2.064

25 1.316 1.708 2.060

26 1.315 1.706 2.056

Tab. 2: Sichel’s factor γn(V) for estimation of the mean of a lognormal population.

Tab. 3: Sichels factor ψ.05(V;n) for lower 95% confidence level on the mean of a lognormal

population.

Tab. 4: Sichels factor ψ.95(V;n) for upper 95% confidence level on the mean of a lognormal

population.