Explore a world of knowledge and get your questions answered on IDNLearn.com. Find the solutions you need quickly and accurately with help from our knowledgeable community.
Sagot :
Sure, let's tackle each part of the problem step-by-step.
Given:
- The probability [tex]\( p \)[/tex] of a Nigerian home having a fridge is 0.35 (or 35%).
- The sample size [tex]\( n \)[/tex] is 20 homes.
1. Find the probability that [tex]\( X \)[/tex] is equal to 5:
- To find this probability, we use the binomial probability mass function (PMF). The binomial PMF formula is:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \][/tex]
where [tex]\( \binom{n}{x} \)[/tex] is the binomial coefficient, representing the number of ways to choose [tex]\( x \)[/tex] successes (homes with refrigerators) out of [tex]\( n \)[/tex] trials (sample of homes).
- For [tex]\( X = 5 \)[/tex]:
[tex]\[ P(X = 5) \approx 0.1272 \][/tex]
2. Find the probability that [tex]\( X \)[/tex] is at least 10:
- To find this probability, we need to consider the cumulative probability for all values from 10 to 20.
- This can be calculated using the cumulative distribution function (CDF) of the binomial distribution:
[tex]\[ P(X \geq 10) = 1 - P(X < 10) \][/tex]
- This means we subtract the cumulative probability of [tex]\( X \)[/tex] being less than 10 (i.e., sum of probabilities from [tex]\( X = 0 \)[/tex] to [tex]\( X = 9 \)[/tex]) from 1:
[tex]\[ P(X \geq 10) \approx 0.1218 \][/tex]
3. Find the probability that [tex]\( X \)[/tex] is at most 9:
- This is directly given by the cumulative distribution function (CDF) for [tex]\( X \leq 9 \)[/tex]:
[tex]\[ P(X \leq 9) \approx 0.8782 \][/tex]
4. Give the mean, variance, and standard deviation of [tex]\( X \)[/tex]:
- The mean of a binomial distribution [tex]\( X \)[/tex] with parameters [tex]\( n \)[/tex] and [tex]\( p \)[/tex] is given by:
[tex]\[ \mu = n \cdot p \][/tex]
So,
[tex]\[ \mu = 20 \cdot 0.35 = 7 \][/tex]
- The variance of [tex]\( X \)[/tex] is given by:
[tex]\[ \sigma^2 = n \cdot p \cdot (1 - p) \][/tex]
So,
[tex]\[ \sigma^2 = 20 \cdot 0.35 \cdot (1 - 0.35) = 20 \cdot 0.35 \cdot 0.65 = 4.55 \][/tex]
- The standard deviation is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \approx \sqrt{4.55} \approx 2.133 \][/tex]
In summary:
- The probability that [tex]\( X = 5 \)[/tex] is approximately 0.1272.
- The probability that [tex]\( X \geq 10 \)[/tex] is approximately 0.1218.
- The probability that [tex]\( X \leq 9 \)[/tex] is approximately 0.8782.
- The mean of [tex]\( X \)[/tex] is 7.
- The variance of [tex]\( X \)[/tex] is 4.55.
- The standard deviation of [tex]\( X \)[/tex] is approximately 2.133.
Given:
- The probability [tex]\( p \)[/tex] of a Nigerian home having a fridge is 0.35 (or 35%).
- The sample size [tex]\( n \)[/tex] is 20 homes.
1. Find the probability that [tex]\( X \)[/tex] is equal to 5:
- To find this probability, we use the binomial probability mass function (PMF). The binomial PMF formula is:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \][/tex]
where [tex]\( \binom{n}{x} \)[/tex] is the binomial coefficient, representing the number of ways to choose [tex]\( x \)[/tex] successes (homes with refrigerators) out of [tex]\( n \)[/tex] trials (sample of homes).
- For [tex]\( X = 5 \)[/tex]:
[tex]\[ P(X = 5) \approx 0.1272 \][/tex]
2. Find the probability that [tex]\( X \)[/tex] is at least 10:
- To find this probability, we need to consider the cumulative probability for all values from 10 to 20.
- This can be calculated using the cumulative distribution function (CDF) of the binomial distribution:
[tex]\[ P(X \geq 10) = 1 - P(X < 10) \][/tex]
- This means we subtract the cumulative probability of [tex]\( X \)[/tex] being less than 10 (i.e., sum of probabilities from [tex]\( X = 0 \)[/tex] to [tex]\( X = 9 \)[/tex]) from 1:
[tex]\[ P(X \geq 10) \approx 0.1218 \][/tex]
3. Find the probability that [tex]\( X \)[/tex] is at most 9:
- This is directly given by the cumulative distribution function (CDF) for [tex]\( X \leq 9 \)[/tex]:
[tex]\[ P(X \leq 9) \approx 0.8782 \][/tex]
4. Give the mean, variance, and standard deviation of [tex]\( X \)[/tex]:
- The mean of a binomial distribution [tex]\( X \)[/tex] with parameters [tex]\( n \)[/tex] and [tex]\( p \)[/tex] is given by:
[tex]\[ \mu = n \cdot p \][/tex]
So,
[tex]\[ \mu = 20 \cdot 0.35 = 7 \][/tex]
- The variance of [tex]\( X \)[/tex] is given by:
[tex]\[ \sigma^2 = n \cdot p \cdot (1 - p) \][/tex]
So,
[tex]\[ \sigma^2 = 20 \cdot 0.35 \cdot (1 - 0.35) = 20 \cdot 0.35 \cdot 0.65 = 4.55 \][/tex]
- The standard deviation is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \approx \sqrt{4.55} \approx 2.133 \][/tex]
In summary:
- The probability that [tex]\( X = 5 \)[/tex] is approximately 0.1272.
- The probability that [tex]\( X \geq 10 \)[/tex] is approximately 0.1218.
- The probability that [tex]\( X \leq 9 \)[/tex] is approximately 0.8782.
- The mean of [tex]\( X \)[/tex] is 7.
- The variance of [tex]\( X \)[/tex] is 4.55.
- The standard deviation of [tex]\( X \)[/tex] is approximately 2.133.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.
