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 1.4 mm. For a randomly found shared, 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) 0.0869

(b) 0.0668

(c) 0.8463

Explanation:

We are given that the 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 1.4 mm.

Let X = thickness measurements of ancient prehistoric Native American pot shards

The z-score probability distribution is given by ;

Z =
`( X-\mu)/(\sigma)} }` ~ N(0,1)

where,
`\mu` = mean thickness = 4.9 mm

`\sigma` = standard deviation = 1.4 mm

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

(a) Probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

P(X < 3.0 mm) = P(
`( X-\mu)/(\sigma)} }` <
`( 3.0-4.9)/(1.4)} }` ) = P(Z < -1.36) = 1 - P(Z
`\leq` 1.36)

= 1 - 0.9131 = 0.0869

The above probability is calculated using z table by looking at value of x = 1.36 in the z table which have an area of 0.9131.

(b) Probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

P(X > 7.0 mm) = P(
`( X-\mu)/(\sigma)} }` >
`( 7.0-4.9)/(1.4)} }` ) = P(Z > 1.50) = 1 - P(Z
`\leq` 1.50)

= 1 - 0.9332 = 0.0668

The above probability is calculated using z table by looking at value of x = 1.50 in the z table which have an area of 0.9332.

(c) Probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X
`\leq` 3.0 mm)

P(X < 7.0 mm) = P(
`( X-\mu)/(\sigma)} }` <
`( 7.0-4.9)/(1.4)} }` ) = P(Z < 1.50) = 0.9332 {using z table}

P(X
`\leq` 3.0 mm) = P(
`( X-\mu)/(\sigma)} }`
`\leq`
`( 3.0-4.9)/(1.4)} }` ) = P(Z
`\leq` -1.36) = 1 - P(Z < 1.36)

= 1 - 0.9131 = 0.0869

Therefore, P(3.0 mm < X < 7.0 mm) = 0.9332 - 0.0869 = 0.8463

Answer 2

a) 0.0874

b) 0.0668

c) 0.8458

Explanation:

We are given the following information in the question:

Mean, μ = 4.9 mm

Standard Deviation, σ = 1.4 mm

We are given that the distribution of thickness is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P( thickness is less than 3.0 mm)

`P( x < 3.0) = P( z < \displaystyle(3.0 - 4.9)/(1.4)) = P(z < -1.3571)`

Calculation the value from standard normal z table, we have,

`P(x < 3.0) = 0.0874`

b) P(thickness is more than 7.0 mm)

`P( x > 7.0) = P( z < \displaystyle(7.0 - 4.9)/(1.4)) = P(z > 1.5)`

`= 1 - P(z \leq 1.5)`

Calculation the value from standard normal z table, we have,

`P(x > 7.0) = 1 - 0.9332 = 0.0668`

c) P(thickness is between 3.0 mm and 7.0 mm)

`P(3.0 < x < 7.0)\\=1 - P(x<3.0) - P(x > 7.0)\\=1 - 0.0874- 0.0668\\=0.8458`