Question

The quality assurance engineer of televisions inspects TV's from a large factory. She randomly selects a sample of 18 TV's from the factory to inspect. Assume that 33% of the TV's in the lot are silver. Let Y be the number of silver TV's selected.

A) find the probability that 2 of the TV's selected are silver.

B) find the probability that less than 4 of the TV's selected are silver.

C) find the probability that between 3.5 and 7.5 of the TV's selected are silver.

D) find the mean of Y.

E) find the standard deviation of Y.

All decimals to the ten-thousandths place please.

Answer 1

This is a binomial event where the two outcomes are either silver or not silver. The probability of every T.V being silver is same.

n = 18 p = 33% = 0.33

(n = 18 ,p = 0.33 )

P( X =x) =

=

WE then sub x = 0 ......18 and calculate probability at each 'x'.

X	P(X=x )
0	0.00074
1	0.00656
2	0.02747
3	0.07217
4	0.13330
5	0.18383
6	0.19618
7	0.16564
8	0.11218
9	0.06139
10	0.02721
11	0.00975
12	0.00280
13	0.00064
14	0.00011
15	0.00001
16	0.00000
17	0.00000
18	0.00000
Total	1.00000

A) find the probability that 2 of the TV's selected are silver.

P(X = 2) =

Ans: 0.0275

B) find the probability that less than 4 of the TV's selected are silver.

P( X < 4) = P(X = 0 ) + P(X = 1 ) + P(X =2 ) + P(X = 3)

= 0.0007 + 0.0066 + 0.0275 + 0.0722

Ans: 0.1069

C) find the probability that between 3.5 and 7.5 of the TV's selected are silver.

Binomial is a discrete event where x = 0,1,2 and no fractional therefore within the range we take all the discrete values

P(3.5 < X < 7.5) = P(X = 4 ) + P(X = 5 ) + P(X =6 ) + P(X = 7)

Ans: 0.6790

D) find the mean of Y.

Expected of binomial = np

Therefore E(Y) = 18 * 0.33

E(Y) = 5.94

E) find the standard deviation of Y.

Var of binomial = np(1-p)

= 18 * 0.33 * 0.67

= 3.9798

SD =

SD(Y) = 1.99494