Question

Use the probability distribution to complete parts​ (a) and​ (b) below. The number of defects per 1000 machine parts inspected Defects 0 1 2 3 4 5

Probability 0.262 0.291 0.240 0.151 0.037 0.019

​(a) Find the​ mean, variance, and standard deviation of the probability distribution.

The mean is

nothing.

​(Round to one decimal place as​ needed.)

The variance is

nothing.

​(Round to one decimal place as​ needed.)

The standard deviation is

nothing.

​(Round to one decimal place as​ needed.)

​(b) Interpret the results.

The mean is

nothing​,

so the average batch of 1000 machine parts has

▼

at least 2 defects.

no defects.

1 or 2 defects.

The standard deviation is

nothing​,

so most of the batches of 1000

▼

differ from the mean by no more than about 1 defect.

do not differ from the mean.

differ from the mean by no more than about 2 defects.

differ from the mean by more than about 2 defects.

​(Round to one decimal place as​ needed.)

Answer 1

(a)

x              p                    xp                  x²p

0             0.262                0                  0

1           0.291                0.291            0.291

2          0.240                 0.480            0.960

3          0.151                  0.453            1.359

4         0.037                   0.148          0.592

5        0.019                  0.095            0.475

--------------------------------------------------------------------

Total                           1.467              3.677

(i) The mean is:

1.5

(ii) Variance = E(X²) - (E(X))² = 3.677 - 1.467² = 1.5249

So,

The variance is:

1.5

(iii) Standard deviation =

So,

The standard deviation is:

1.2

(b)

(i)

The mean is 1.5 , so the average batch of 1000 machine parts has 1 or 2 defectives.

(ii)

The standard deviation is 1.2, so most of the batches of 1000 differ from the mean by no more than about 2 defects.