Higher Engineering Mathematics

584 STATISTICS AND PROBABILITY

Problem 6. The mean diameter of a long length

of wire is to be determined. The diameter of the

wire is measured in 25 places selected at ran-

dom throughout its length and the mean of these

values is 0.425 mm. If the standard deviation of

the diameter of the wire is given by the manu-

facturers as 0.030 mm, determine (a) the 80%

confidence interval of the estimated mean diam-

eter of the wire, and (b) with what degree of

confidence it can be said that ‘the mean diameter

is 0.425±0.012 mm’.

For the population:σ= 0 .030 mm
For the sample:N=25,x= 0 .425 mm
Since an infinite number of measurements can
be obtained for the diameter of the wire, the pop-
ulation is infinite and the estimated value of the
confidence interval of the population mean is given
by expression (5).

(a) For an 80% confidence level, the value ofzcis
obtained from Table 61.1 and is 1.28.
The 80% confidence level estimate of the confi-
dence interval of

μ=x±

zcσ

√

N

= 0. 425 ±

(1.28)(0.030)

√

25

= 0. 425 ± 0 .0077 mm

i.e. the 80% confidence interval is from
0.417 mm to 0.433 mm.
This indicates that the estimated mean diameter
of the wire is between 0.417 mm and 0.433 mm
and that this prediction is likely to be correct
80 times out of 100.
(b) To determine the confidence level, the given data
is equated to expression (5), giving

0. 425 ± 0. 012 =x±zc

σ

√

N

Butx=0.425 therefore

±zc

σ

√

N

=± 0. 012

i.e.zc=

0. 012

√

N

σ

=±

(0.012)(5)

0. 030

=± 2

Using Table 58.1 of partial areas under the

standardized normal curve, a zc value of 2

standard deviations corresponds to an area of

0.4772 between the mean value (zc=0) and+ 2

standard deviations. Because the standardized

normal curve is symmetrical, the area between

the mean and±2 standard deviations is

0.4772×2, i.e. 0.9544.

Thus the confidence level corresponding to

0. 425 ± 0 .012 mm is 95.44%.

(b) Estimating the mean and standard deviation of

a population from sample data

The standard deviation of a large population is not

known and, in this case, several samples are drawn

from the population. The mean of the sampling dis-

tribution of means,μxand the standard deviation of

the sampling distribution of means (i.e. the standard

error of the means),σx, may be determined. The con-

fidence limits of the mean value of the population,

μ, are given by

μx±zcσx (6)

wherezcis the confidence coefficient corresponding

to the confidence level required.

To make an estimate of the standard deviation,σ,

of a normally distributed population:

(i) a sampling distribution of the standard devia-

tions of the samples is formed, and

(ii) the standard deviation of the sampling distribu-

tion is determined by using the basic standard

deviation formula.

This standard deviation is called the standard error

of the standard deviations and is usually signified by

σs.Ifsis the standard deviation of a sample, then

the confidence limits of the standard deviation of the

population are given by:

s±zcσs (7)

wherezcis the confidence coefficient corresponding

to the required confidence level.

Problem 7. Several samples of 50 fuses

selected at random from a large batch are tested

when operating at a 10% overload current and

the mean time of the sampling distribution

before the fuses failed is 16.50 minutes. The

standard error of the means is 1.4 minutes. Deter-

mine the estimated mean time to failure of the

batch of fuses for a confidence level of 90%.

For the sampling distribution: the mean,μx= 16 .50,

the standard error of the means,σx= 1 .4.

The estimated mean of the population is based on

sampling distribution data only and so expression