Question

Magnetic surveying is one technique used by archaeologists to determine anomalies arising from variations in magnetic susceptibility. Unusual changes in magnetic susceptibility might (or might not) indicate an important archaeological discovery. Let x be a random variable that represents a magnetic susceptibility (MS) reading for a randomly chosen site at an archaeological research location. A random sample of 120 sites gave the readings shown in the table below.

Magnetic Susceptibility Readings, centimeter-gram-second ✕ 10−6 (cmg ✕ 10−6)
Comment	Magnetic Susceptibility	Number of Readings	Estimated Probability
"cool"	0 ≤ x < 10	30	30/120 = 0.25
"neutral"	10 ≤ x < 20	54	54/120 = 0.45
"warm"	20 ≤ x < 30	18	18/120 = 0.15
"very interesting"	30 ≤ x < 40	12	12/120 = 0.10
"hot spot"	40 ≤ x	6	6/120 = 0.05

Suppose a reading between 30 and 40 is called "very interesting" from an archaeological point of view. Let us say you take readings at

n = 12

sites chosen at random. Let r be a binomial random variable that represents the number of "very interesting" readings from these 12 sites.

(a) Let us call "very interesting" a binomial success. Use the table above to find p, the probability of success on a single trial, where

p = P(success) = P(30 ≤ x < 40).

(b) What is the expected value μ and standard deviation σ for the random variable r? (Round your standard deviation to three decimal places.)

μ =
σ =

(c) What is the probability that you will find at least one "very interesting" reading in the 12 sites? (Round your answer to three decimal places.)


(d) What is the probability that you will find fewer than four "very interesting" readings in the 12 sites? (Round your answer to three decimal places.)

Answer 1

(a)

From the given table,

p = P(success) = P(30 ≤ x < 40) = 0.10

(b)

n = 12

and r ~ Binomial(n = 12, p = 0.10)

μ = np = 12 * 0.10 = 1.2

σ = = = 1.039

(c)

Probability that you will find at least one "very interesting" reading in the 12 sites = P(r 1)

= 1 - P(r < 1)

= 1 - P(r = 0)

= 1 - ¹²C₀ * 0.1⁰ * (1 - 0.1)^12-0

= 1 - 0.9¹²

= 0.718

(d)

Probability that you will find fewer than four "very interesting" reading in the 12 sites = P(r < 4)

= P(r = 0) + P(r = 1) + P(r = 2) + P(r = 3)

=  ¹²C₀ * 0.1⁰ * (1 - 0.1)^12-0 + ¹²C₁* 0.1¹ * (1 - 0.1)^12-1 + ¹²C₂ * 0.1² * (1 - 0.1)^12-2 + ¹²C₃ * 0.1³ * (1 - 0.1)^12-3

= 0.28242954 + 0.37657272 + 0.23012777 + 0.08523251

= 0.974