Question

A bottling machine can be regulated so that it discharges an average of μ ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with σ = 1.0 ounce. If a sample of  n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting), and the ounces of fill are measured for each, then the probability that the sample mean will be within 0.29 ounce of the true mean is 0.6157. Suppose that Y is to be computed using a sample of size n.

(a) If n = 16, what is P(|Y − μ| ≤ 0.29)? (Round your answer to four decimal places.)

(b) Find P(|Y − μ| ≤ 0.29) when Y is to be computed using samples of sizes n = 25, n = 36, n = 49, and n = 64. (Round your answers to four decimal places.)

n = 25 P(|Y − μ| ≤ 0.29) =

n = 36. P(|Y − μ| ≤ 0.29) =

n = 49. P(|Y − μ| ≤ 0.29) =

n = 64. P(|Y − μ| ≤ 0.29) =

Answer 1

a)

P(|X̅-µ|< 0.29) =

= P(-0.29 < (X̅-µ) < 0.29)

= P( (-0.29)/(1/√16) < (X-µ)/(σ/√n) < (0.29)/(1/√16) )

= P(-1.16 < z < 1.16)

= P(z < 1.16) - P(z < -1.16)

Using excel function:

= NORM.S.DIST(1.16, 1) - NORM.S.DIST(-1.16, 1)

= 0.7540

b) i)

P(|X̅-µ|< 0.29) =

= P(-0.29 < (X̅-µ) < 0.29)

= P( (-0.29)/(1/√25) < (X-µ)/(σ/√n) < (0.29)/(1/√25) )

= P(-1.45 < z < 1.45)

= P(z < 1.45) - P(z < -1.45)

Using excel function:

= NORM.S.DIST(1.45, 1) - NORM.S.DIST(-1.45, 1)

= 0.8529

ii)

P(|X̅-µ|< 0.29) =

= P(-0.29 < (X̅-µ) < 0.29)

= P( (-0.29)/(1/√36) < (X-µ)/(σ/√n) < (0.29)/(1/√36) )

= P(-1.74 < z < 1.74)

= P(z < 1.74) - P(z < -1.74)

Using excel function:

= NORM.S.DIST(1.74, 1) - NORM.S.DIST(-1.74, 1)

= 0.9181

iii)

P(|X̅-µ|< 0.29) =

= P(-0.29 < (X̅-µ) < 0.29)

= P( (-0.29)/(1/√49) < (X-µ)/(σ/√n) < (0.29)/(1/√49) )

= P(-2.03 < z < 2.03)

= P(z < 2.03) - P(z < -2.03)

Using excel function:

= NORM.S.DIST(2.03, 1) - NORM.S.DIST(-2.03, 1)

= 0.9576

iv)

P(|X̅-µ|< 0.29) =

= P(-0.29 < (X̅-µ) < 0.29)

= P( (-0.29)/(1/√64) < (X-µ)/(σ/√n) < (0.29)/(1/√64) )

= P(-2.32 < z < 2.32)

= P(z < 2.32) - P(z < -2.32)

Using excel function:

= NORM.S.DIST(2.32, 1) - NORM.S.DIST(-2.32, 1)

= 0.9797