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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.
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