Higher Engineering Mathematics

580 STATISTICS AND PROBABILITY

given by

σx=

σ

√

N

√(

Np−N

Np− 1

)

In addition, the sample distribution would have

been approximately normal. Assume that the sample

given in the problem is one of many samples. For

many (theoretical) samples:

the mean of the sampling distribution of means,

μx=μ= 6 .5 kg.

Also, the standard error of the means,

σx=

σ

√

N

√(

Np−N

Np− 1

)

=

0. 5

√

60

√(

1500 − 60

1500 − 1

)

= 0 .0633 kg

Thus, the sample under consideration is part of a nor-
mal distribution of mean value 6.5 kg and a standard
error of the means of 0.0633 kg.
(a) If the combined mass of 60 ingots is between
378 and 396 kg, then the mean mass of each of

the 60 ingots lies between

378

60

and

396

60

kg, i.e.

between 6.3 kg and 6.6 kg.

Since the masses are normally distributed, it is

possible to use the techniques of the normal

distribution to determine the probability of the

mean mass lying between 6.3 and 6.6 kg. The

normal standard variate value,z,isgivenby

z=

x−x

σ

,

hence for the sampling distribution of means,

this becomes,

z=

x−μx

σx

Thus, 6.3 kg corresponds to a z-value of

6. 3 − 6. 5

0. 0633

=− 3 .16 standard deviations.

Similarly, 6.6 kg corresponds to az-value of

6. 6 − 6. 5

0. 0633

= 1 .58 standard deviations.

Using Table 58.1 (page 561), the areas corre-

sponding to these values of standard deviations

are 0.4992 and 0.4430 respectively. Hencethe

probability of the mean mass lying between

6.3 kg and 6.6 kg is 0. 4992 + 0. 4430 = 0. 9422.

(This means that if 10 000 samples are drawn,

9422 of these samples will have a combined

mass of between 378 and 396 kg.)

(b) If the combined mass of 60 ingots is 399 kg, the

mean mass of each ingot is

399

60

, that is, 6.65 kg.

The z-value for 6.65 kg is

6. 65 − 6. 5

0. 0633

, i.e.

2.37 standard deviations. From Table 58.1

(page 561), the area corresponding to thisz-

value is 0.4911. But this is the area between the

ordinatez=0 and ordinatez= 2 .37. The ‘more

than’ value required is the total area to the right

of thez=0 ordinate, less the value between

z=0 andz= 2 .37, i.e. 0. 5000 − 0 .4911. Thus,

since areas are proportional to probabilities for

the standardized normal curve,the probability

of the mean mass being more than 6.65 kg

is 0. 5000 − 0 .4911, i.e.0.0089. (This means

that only 89 samples in 10000, for example, will

have a combined mass exceeding 399 kg.)

Now try the following exercise.

Exercise 220 Further problems on the

sampling distribution of means

1. The lengths of 1500 bolts are normally dis-
   tributed with a mean of 22.4 cm and a stan-
   dard deviation of 0.0438 cm. If 30 samples
   are drawn at random from this population,
   each sample being 36 bolts, determine the
   mean of the sampling distribution and stan-
   dard error of the means when sampling is
   done with replacement.

[μx= 22 .4 cm,σx= 0 .0080 cm]

2. Determine the standard error of the means
   in Problem 1, if sampling is done without
   replacement, correct to four decimal places.

[σx= 0 .0079 cm]

3. A power punch produces 1800 washers per
   hour. The mean inside diameter of the wash-
   ers is 1.70 cm and the standard deviation is
   0.013 cm. Random samples of 20 washers are
   drawn every 5 minutes. Determine the mean
   of the sampling distribution of means and the
   standard error of the means for the one hour’s
   output from the punch, (a) with replacement