...,........................

The quality control officer at a manufacturing plant needs to determine what percentage of items in a batch are defective. The f

artcher [175]

Answer:

A. Simple Random Sampling

B. Cluster Sampling

C. Convenience Sampling

D. Systematic Sampling

E. Stratified Sampling

Step-by-step explanation:

Simple Random Sampling is the sampling when samples are chosen randomly, where each unit has an equal chance of being selected in a sample.

If the population is divided into a different group called cluster and clusters are selected as a sample then it is Cluster Sampling.

In Convenience sampling, observers collect the sample as his\her convenience.

In Systematic Sampling sample is chosen by some criteria like he\she is taken every 10th unit as a sample from the population.

In Stratified Sampling population is divided into several groups such that within the group it is homogeneous and between the group it is heterogeneous. And now a selection of each stratum and unit has an equal chance of selection.

8 0

4 years ago

Savannah practice ballet in 1.67 hours on Monday on Wednesday she practice for 3.25 how much longer did savannah practice on Wed

xz_007 [3.2K]

1.58 hours or about an hour and a half. hope this helps :)

5 0

3 years ago

in a certain year the average e-mail user sent or received 148 e-mails each day. The number of e-mails received was 4 more than

Lorico [155]

Answer:

1136 received each day

Step-by-step explanation:

148(2)+4= 1136

8 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

What is <br> 2n - 3 = 9 ( solve the equation)

BartSMP [9]

$2n - 3 = 9 \\ 2n = 12 \\ n = 6$

5 0

4 years ago