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Based on a research survey, when 1020 adults were asked about hand hygiene, 44% said that they wash their hands after using pu

aleksley [76]

Considering that the subjects are chosen without replacement, they are not independent, and the probability cannot be found using the binomial distribution.

The binomial distribution and the hypergeometric distribution are quite similar, as:

They find the probability of exactly x successes on n repeated trials.
For each trial, there are only two possible outcomes.

The difference is that the binomial distribution is for independent trials, that is, in each trial, the probability of success is the same, while the hypergeometric distribution is for dependent trials.

If the sample is without replacement, the trials are not independent, thus the hypergeometric distribution is used, not the binomial.

A similar problem is given at brainly.com/question/21772486

7 0

1 year ago

A video company randomly selected 100 of its subscribers and asked them how many hours of shows they watch per week. Of those su

Xelga [282]

Hello there!

35/100 watch more then 10 hours a week

3000 is 30 times greater than 100 so we can simply multiply 35/100 by 30.

35/100 x 30 = 1050/3000

Therefore, if the company had 3,000 subscribers, 1,050 subscribers would be watching more than 10 hours a week.

6 0

2 years ago

If there are approximately 1.8 million cars in Vancouver and if each car travels an average of 17,500 kilometres in a year, how

Goryan [66]

Rates are used to measure a quantity over another.

The 1.8 million cars use $1.85 \times 10^{10}$ liters each year

Given

$Cars = 1.8\ million$

$d = 17500km$ --- distance

$r = 40miles/gallon$

First, we calculate the number (n) of gallons used by each car

$r = \frac{d}{n}$

Solve for n

$n = \frac{d}{r}$

So, we have:

$n = \frac{17500km}{40miles/gallon}$

$n = \frac{17500km}{40miles}gallon$

Convert miles to kilometers

$n = \frac{17500km}{40km \times 1.60934}gallon$

$n = \frac{17500km}{64.3738km}gallon$

$n = 271.85\ gallon$

The number of gallons (N), used by all the cars is:

$N = n \times cars$

$N = 271.85 \times 1.8\ million$

$N = 489330000$

Convert to liters

$N = 489330000 \times 3.78541L$

$N = 1852314675.3L$

In scientific notation to 2 decimal places, we have:

$N = 1.85 \times 10^{10}L$

Hence, the number of liters used is $1.85 \times 10^{10}$

Read more about distance and rates at:

brainly.com/question/24659604

4 0

2 years ago

In a set of 25 aluminum castings, four castings are defective (D), and the remaining twenty-one are good (G). A quality control

lianna [129]

Answer:

The sample space for selecting the group to test contains 2,300 elementary events.

Step-by-step explanation:

There are a total of N = 25 aluminum castings.

Of these 25 aluminum castings, n₁ = 4 castings are defective (D) and n₂ = 21 are good (G).

It is provided that a quality control inspector randomly selects three of the twenty-five castings without replacement to test.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!(n-k)!}$

Compute the number of samples that are possible as follows:

${25\choose 3}=\frac{25!}{3!\times (25-3)!}$

$=\frac{25\times 24\times 23\times 22!}{3!\times 22!}\\\\=\frac{25\times 24\times 23}{3\times 2\times 1}\\\\=2300$

The sample space for selecting the group to test contains 2,300 elementary events.

6 0

2 years ago

g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa

adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is $\sigma = 0.0133$