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RSB [31]
3 years ago
7

In the school election James ran against Ben for secretary James received 75% of the votes if 240 students voted how many votes

did James not receive?
Mathematics
1 answer:
scoray [572]3 years ago
3 0

Answer: James did not received 60 votes.

Step-by-step explanation:

Given: In a school election,

Percentage of votes received by James = 75%

Percentage of voted hadn't received by James = 100% - 75% = 25%

If 240 students voted , the number of votes hadn't received by James= 25% of 240

=\dfrac{25}{100}\times 240\\\\=\dfrac14\times240\\\\=60

Hence, James did not received 60 votes.

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Twenty-five samples of 100 items each were inspected when a process was considered to be operating satisfactorily. In the 25 sam
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a) The estimate of the proportion of defectives when the process is in control is 0.054

b) The standard error of the proportion if the sample size is 100 is 0.0226.

c) The upper control limit is 0.1218 and the lower control limit is 0 (since LCL < 0 and p > 0, we can write LCL = 0).

<h3>What are the formulas for finding the estimate of the proportion, standard variation, and control limits?</h3>

1) The estimate of the proportion of success is

p = (number of success)/(total number of samples)

I.e., p = x/N

2) The standard deviation of the proportion of success is

\sigma_p = \sqrt{\frac{p(1-p)}{n} }

3) The upper and lower control limits for a control chart are:

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<h3>Calculation:</h3>

It is given that, there are 25 samples of 100 items each.

So, the total number of items i.e., the total sample size,

N = 25 × 100 = 2500

In 25 samples, a total of 135 items were found to be defective.

So, the number of defectives x = 135

a) The estimate of the proportion of defectives is p = x/N

On substituting, we get

p = 135/2500 = 0.054

b) The standard error of the proportion if the sample of size 100 is calculated by

\sigma_p = \sqrt{\frac{p(1-p)}{n} }

On substituting p = 0.054 and  n = 100, we get

\sigma_p = \sqrt{\frac{0.054(1-0.54)}{100} }

    = 0.0226

c) The control limits for the control chart are:

Upper control limit =  p + 3\sigma_p

⇒ U.C.L = 0.054 + 3(0.0226) = 0.054 + 0.0678 = 0.1218

Lower control limit = p - 3\sigma_p

⇒ L.C.L = 0.054 - 3(0.0226) = 0.054 - 0.0678 = - 0.0138 ≈ 0

(Since we know that the lower control limit should not be a negative value, it is made equal to 0).

Learn more about an estimate of the proportion here:

brainly.com/question/23986522

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