Question

A lab machine fills test tubes at a mean volume of 9 ml with a standard deviation of 0.35 ml. If we continually sample 49 test tubes this machine has filled and for each sample, we collect, compute x ̅, what proportion of the samples would you expect to be… a)below 8.92ml? b)above 9.08ml? c)above 9ml? d)Fill in the blank. ̅x greater than ________ml is unlikely. Only 5% or 1 out of 20 samples would yield an x ̅ greater than this volume.

Answer 1

Solution

Let X = volume (in ml) filled by the lab machine.

We assume X ~ N(µ, σ²), where given µ = 9, and σ = 0.35 …………………………………… (1)

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…………........……...…(2)

X bar ~ N(µ, σ²/n),…………………………………………………………......................….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar ≤ or ≥ t) = P[Z ≤ or ≥ {(√n)(t - µ)/σ}] ……………………………....................……(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables …………………………………………………..…….....…………… (5a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) …(5b)

Percentage points of N(0, 1) can be found using Excel Function: Statistical, NORMSINV

which gives values of t for which P(Z ≤ t) = given probability….…. ………...……………..…(5c)

Now to work out the solution,

Part (a)

Proportion of the samples that is expected below 8.92 ml

= P(X < 8.92)

= P[Z < {(8.92 - 9)/0.35}] [vide (2) and (1)]

= P[Z < - 0.2285]

= 0.4096 [vide (5b)] Answer 1

Part (b)

Proportion of the samples that is expected above 9.08 ml

= P(X > 9.08)

= P[Z > {(9.08 - 9)/0.35}] [vide (2) and (1)]

= P[Z > 0.2285]

= 0.4096 [vide (5b)] Answer 2

Part (c)

Proportion of the samples that is expected above 9 ml

= P(X > 9)

= P[Z > {(9 - 9)/0.35}] [vide (2) and (1)]

= P[Z > 0]

= 0.5 [vide (5b)] Answer 3

Part (d)

We want a value t of Xbar such that P(Xbar > t) = 0.05

i.e., vide vide (4) and (1), P[Z > {(√49)(t - 9)/0.35}] = 0.05

=> vide (5c), 7(t - 9)/0.35 = 1.645

Or, t = 9.08.

Thus, ̅x greater than 9.08 ml is unlikely Answer 4

DONE

Going Beyond,

Answers 2 and 3 could be directly written using the

Property of symmetry: P(Z ≥ t) = P(Z ≤ - t)(7)

P(Z ≤ 0) = P(Z ≥ 0 ) = ½

Complete