Answer to Question #292284 in Statistics and Probability for Yaku

Question

8. A quality control engineer wants to check whether, in accordance with specication, 90% of products shipped are in perfect working condition. To this end, she randomly selects 12 items from each lot ready to be shipped and passes the lot only if all 12 are in perfect working condition. If one or more are not in perfect working condition, she holds the lot for a complete inspection. Find the probability that she will commit the error of

a) holding a lot for a complete inspection even though 90% of the items are in perfect working condition;

b) letting a lot pass through even though only 80% of the items are in perfect working condition;

10. Find the area under the standard normal curve that lies

a) between z = 0 and z = 0.87;

b) between z = -1.66 and z = 0;

c) to the right of z = 0.48;

d) to the right of z = -0.27;

e) to the left of z = 1.30;

f) between z = 0.45 and z = 1.23;

h) between z = 1.15 and z = 1.23;

i) between z = -1.35 and z = 1.35

j) between z = -2.35 and z = -1.46.

Answer 1

8)

a) Denote X as the number of defectives found

From this we may say X belongs to a binomial distribution with n = 12 and p = 0.1

Now, the probability that X is greater than or equal to one = 1 - P( X = 0 )

= ( 1 - ( 0.1)¹²)

= 1

b) Denote X as the number of non defectives found

assuming X belongs to a binomial distribution with n =12 and p = 0.8

Hence P(X = 12) = (0.8)¹²

= 0.068719476

10)

a)  between z = 0 and z = 0.87

we define P( 0 < z < 0.87)

using the normal table, we have

( 0.80785 - 0.5 ) = 0.30785 which is the required area

b) between z = -1.66 and z = 0;

we define P( -1.66 < z < 0)

using the normal table we have

(0.50000 - 0.04846) = 0.45154 which is the required area

c)  to the right of z = 0.48;

we define 1 - P( z = 0.48 )

using the normal table, we have

= 1 - ( 0.68439)

= 0.31561 which is the required area

d) to the right of z = -0.27;

we define 1- P(z = - 0.27)

using the normal table, we have

= 1 - (0.39358)

= 0.60642 which is the required area

e) to the left of z = 1.30

Define P( z = 1.30)

using the normal table, we

= 0.90320 which is the required area

f)  between z = 0.45 and z = 1.23;

Define P( 0.45 < z < 1.23)

using the normal table, we have

= ( 0.89065 - 0.67364)

= 0.21701 which is the required area

h) between z = 1.15 and z = 1.23;

Define P( 1.15 < z < 1.23 )

using the normal table we have

= ( 0.89065 - 0.87493)

= 0.01572 which is the required area

i) between z = -1.35 and z = 1.35

Define P( -1.35 < z < 1.35)

using the normal table, we have

= ( 0.91149 - 0.08851)

= 0.82298 which is the required area

j)  between z = -2.35 and z = -1.46.

Define P( -2.35 < z < -1.46)

using the normal table, we have

= ( 0.07215 - 0.00939)

= 0.06276 which is the required area.