Question

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.9 millimeters (mm) and a standard deviation of 1.2 mm. For a randomly found shard, find the following probabilities. (Round your answers to four decimal places.)

(a) the thickness is less than 3.0 mm
(b) the thickness is more than 7.0 mm
(c) the thickness is between 3.0 mm and 7.0 mm

Answer 1

a) The probability that the thickness is less than 3.0 mm

P(X<3.0) = 0.057

b) The probability that the thickness is less than 7.0 mm

P(X>7) = 0.0401

c) The probability that the thickness is between 3.0 mm and 7.0 mm

P( 3 < x < 7)= 0.9029

Explanation:

Step(i):-

Given mean of the Population (μ) = 4.9 mm

Standard deviation of the Population(σ) = 1.2 mm

a)

Let 'X' be the random variable in normal distribution

`Z = (x-mean)/(S.D) = (3.0-4.9)/(1.2) = -1.583`

The probability that the thickness is less than 3.0 mm

`P(X<3.0) = P(Z < -1.583) = 1 - P( Z>1.583)`

= 1 - ( 0.5 + A(1.583)

= 0.5 - A(1.583)

= 0.5 - 0.4430

= 0.057

The probability that the thickness is less than 3.0 mm

P(X<3.0) = 0.057

b)

Let 'X' be the random variable in normal distribution

`Z = (x-mean)/(S.D) = (7.0-4.9)/(1.2) = 1.75`

The probability that the thickness is less than 7.0 mm

`P(X>7.0) = P(Z >1.75) = 0.5 - A(1.75)`

= 0.5 - 0.4599 ( from normal table )

= 0.0401

The probability that the thickness is less than 7.0 mm

P(X>7) = 0.0401

c)

Let 'X' be the random variable in normal distribution

`Z_(1) = (x_(1) -mean)/(S.D) = (3.0-4.9)/(1.2) = -1.583`

`Z_(2) = (x_(2) -mean)/(S.D) = (7.0-4.9)/(1.2) = 1.75`

The probability that the thickness is between 3.0 mm and 7.0 mm

P( 3 < x < 7) = P( - 1.583 < X < 1.75 )

= A( 1.75 ) + A( -1.583)

= A(1.75)+A(1.583)

= 0.4599 + 0.4430

= 0.9029

The probability that the thickness is between 3.0 mm and 7.0 mm

P( 3 < x < 7)= 0.9029