Question

A geologist has collected 14 specimens of basaltic rock and 14 specimens of granite. The geologist instructs a laboratory assistant to randomly select 23 of the specimens for analysis.

(a)

What is the pmf of the number of granite specimens selected for analysis? (Round your probabilities to four decimal places.)

x
p(x)

(b)

What is the probability that all specimens of one of the two types of rock are selected for analysis? (Round your answer to four decimal places.)

(c)

What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value? (Round your answer to four decimal places.)

Answer 1

Solution:

We are given that: a geologist has collected 14 specimens of basaltic rock and 14 specimens of granite.

Thus total rocks = N = 14 +14 = 28

The geologist instructs a laboratory assistant to randomly select 23 of the specimens for analysis.

Thus sample size = n = 23

Part a) What is the pmf of the number of granite specimens selected for analysis?

x = the number of granite specimens selected

Thus M = Number of granite specimens = 14

Thus x can have following values:

x	n-x = 23-x
9	14
10	13
11	12
12	11
13	10
14	9

We can select minimum 9 granite specimens, so that total becomes 9 granite specimens + 14 basaltic rock = 23

and maximum 14 granite specimens, so that total becomes 14 granite specimens + 9 basaltic rock = 23

Here x = the number of granite specimens selected follows Hypergeometric distribution with parameters N = 28 ,M=14 , n = 23 .

Thus using Hypergeometric distribution:

use excel command to find combinations terms:

=COMBIN( n , x )

=COMBIN( 14 , 9)

=2002

=COMBIN(14,14)

=1

and

=COMBIN(28,23)

=98280

Thus

Similarly we find all next probabilities:

x	P(x)
9	0.0204
10	0.1426
11	0.3370
12	0.3370
13	0.1426
14	0.0204

We can use Excel command of Hypergeometric distribution as:

=HYPGEOM.DIST( x , M , n , N , cumulative )

x	Excel command	P(x)
9	= HYPGEOM.DIST( 9 , 14 , 23 , 28 ,FALSE)	0.0204
10	= HYPGEOM.DIST( 10 , 14 , 23 , 28 ,FALSE)	0.1426
11	= HYPGEOM.DIST( 11 , 14 , 23 , 28 ,FALSE)	0.3370
12	= HYPGEOM.DIST( 12 , 14 , 23 , 28 ,FALSE)	0.3370
13	= HYPGEOM.DIST( 13 , 14 , 23 , 28 . FALSE)	0.1426
14	= HYPGEOM.DIST( 14 , 14 , 23 , 28 , FALSE)	0.0204

Part b) What is the probability that all specimens of one of the two types of rock are selected for analysis?

the probability that all specimens of one of the two types of rock are selected

That is:
P( All 14 are granite specimens and rest 9 are basaltic rock OR All 14 are basaltic rock and rest 9 are granite specimens)=......?

That is:
P( 14 granite specimens , 9 basaltic rock OR 14 basaltic rock , 9 granite specimens) =......?

From above table in part a) , we have:

P( 14 granite specimens) = 0.0204 , that means

P(14 granite specimens , 9 basaltic rock) = 0.0204

and

P( 9 granite specimens)=0.0204

That is:
P( 14 basaltic rock , 9 granite specimens) = 0.0204

Thus

P( 14 granite specimens , 9 basaltic rock OR 14 basaltic rock , 9 granite specimens) =0.0204 + 0.0204

P( 14 granite specimens , 9 basaltic rock OR 14 basaltic rock , 9 granite specimens) = 0.0408

Part c) What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?

Find mean and Standard deviation using hypergeometric distribution.

and

Thus standard deviation is:

Thus the number of granite specimens selected for analysis is within 1 standard deviation of its mean value are:

Mean - SD = 11.5 - 1.0319 = 10.4681

Mean + SD = 11.5 + 1.0319 = 12.5319

Thus x values between 10.4681 and 12.5319 are respectively ( 11 , 12 )

Thus

the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value is

=P(X=11)+P(X=12)

=0.3370 +0.3370

=0.6740