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2 years ago

When conducting an experiment, is it better to include 10 students or 30

Mathematics

1 answer:

emmainna [20.7K]2 years ago

8 0

Answer:

Step-by-step explanation:

It is better to include 30 students for the sample. When conducting an experiment it is always best to have the largest sample size possible. The larger the sample size the more data that will be gathered and analyzed, which ultimately leads to much more precise results. Mainly, since you have more results to compare with one another. Therefore, providing a smaller margin of error while also allowing for a better chance of catching errors as opposed to smaller data/sample sizes.

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2 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = the number of defective boards in a random sample of size, n = 25

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= 0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= 0.034

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$