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Plsss Any LAW Student/lawyer That Can Help Me With This Assignment (2) (3) (4)
Please I Need Help With This Assignment by Weezyval(m): 11:32am On May 25, 2015 |
please can anybody help me out here I have an assigment but I don't know how to go about it QUESTION ONE in a survey of 100 students, those studying various languages were found to be: Spanish 28; German 30; French 42; Spanish and French 10; Spanish and German 8; German and French 5; all the three languages 3 (I) how many students studied no language? (ii) how many students had French as their only language? QUESTION TWO in a manufacturing process 20% of items produced are defective. 6 items are chosen from the production line, determine the probability of getting the following (I) exactly one? (ii) none? (iii) more than two? I will be very grate full if any mathematician can help me out pleasssss tanks |
Re: Please I Need Help With This Assignment by chiibekee(f): 11:40am On May 25, 2015 |
Am sorry I can't help. I suck when it comes to mathematics. Would have loved to help. |
Re: Please I Need Help With This Assignment by Weezyval(m): 11:46am On May 25, 2015 |
chiibekee:OK but tanks anyway 1 Like |
Re: Please I Need Help With This Assignment by Lisajames(f): 11:31am On Jan 08 |
Seeking assistance is a wise move. For top-notch support, check out Assignment Desk. Our experts are ready to provide the help you need. Embrace success with our exceptional assignment help services! 1 Like |
Re: Please I Need Help With This Assignment by IreMIDE1: 10:33pm On Jan 08 |
**Question One:** **(I) How many students studied no language?** To find the number of students who studied no language, we can use the principle of inclusion-exclusion. First, add the individual counts of students studying each language: Spanish: 28 German: 30 French: 42 Now, account for those studying more than one language: Spanish and French: 10 Spanish and German: 8 German and French: 5 Also, subtract the count of students studying all three languages: All three languages: 3 Let's calculate: Total students studying at least one language = (Spanish) + (German) + (French) - (Students studying two languages) + (Students studying all three languages) Total students studying at least one language = 28 + 30 + 42 - (10 + 8 + 5) + 3 Total students studying at least one language = 100 Therefore, the number of students studying no language = Total students - Students studying at least one language Number of students studying no language = 100 - 100 Number of students studying no language = 0 So, there are **0 students** who studied no language. **(ii) How many students had French as their only language?** To find the number of students who studied only French, subtract the students studying French and another language(s) from the total number studying French. Students studying only French = French - (French and Spanish) - (French and German) + (All three languages) Let's calculate: Students studying only French = French - (French and Spanish) - (French and German) + (All three languages) Students studying only French = 42 - 10 - 5 + 3 Students studying only French = 30 So, there are **30 students** who had French as their only language. **Question Two:** **(I) Probability of exactly one defective item out of 6 chosen?** This can be calculated using the binomial probability formula: Probability of getting exactly \(k\) successes in \(n\) trials = \(\binom{n}{k} \times p^k \times (1-p)^{n-k}\) Where: \(n\) = total trials (6 items chosen) \(k\) = number of successes (exactly one defective item) \(p\) = probability of success (probability of choosing a defective item = 20% or 0.2) Let's calculate: Probability of exactly one defective item = \(\binom{6}{1} \times 0.2^1 \times (1-0.2)^{6-1}\) Probability of exactly one defective item = \(6 \times 0.2 \times 0.8^5\) Probability of exactly one defective item ≈ \(0.329\) Therefore, the probability of getting exactly one defective item out of 6 chosen is approximately **0.329**. **(ii) Probability of none of the 6 items being defective?** The probability of not choosing a defective item in a single trial is \(1 - 0.2 = 0.8\). To find the probability of none of the 6 items being defective, we raise the probability of not choosing a defective item to the power of the number of trials. Probability of none defective in 6 items = \(0.8^6\) Probability of none defective in 6 items ≈ \(0.262\) Therefore, the probability of none of the 6 items being defective is approximately **0.262**. **(iii) Probability of more than two defective items out of 6 chosen?** To find the probability of more than two defective items, you can find the individual probabilities of getting 0, 1, and 2 defective items, and then subtract their sum from 1 to find the probability of getting more than two defective items. Probability of more than two defective items = 1 - (Probability of 0 defective + Probability of 1 defective + Probability of 2 defective) We've already calculated the probabilities of 0 and 1 defective items. To find the probability of exactly 2 defective items: Probability of exactly two defective items = \(\binom{6}{2} \times 0.2^2 \times (1-0.2)^{6-2}\) Probability of exactly two defective items = \(15 \times 0.04 \times 0.8^4\) Probability of exactly two defective items ≈ \(0.331\) Now, let's calculate the probability of more than two defective items: Probability of more than two defective items = 1 - (Probability of 0 defective + Probability of 1 defective + Probability of 2 defective) Probability of more than two defective items = 1 - (0.262 + 0.329 + 0.331) Probability of more than two defective items ≈ \(0.078\) Therefore, the probability of getting more than two defective items out of 6 chosen is approximately **0.078**. I am an English student, but chatgpt did this. Pls check well ooo, I don't know if it is totally correct |
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