Answer :
Certainly! Let's address each part of the question step-by-step.
### Part i: Mean, Variance, and Standard Deviation of [tex]\( X \)[/tex]
Given:
- Probability of a worker not having health insurance, [tex]\( p = 0.4 \)[/tex].
- Number of workers in the sample, [tex]\( n = 25 \)[/tex].
#### Mean
The mean [tex]\( \mu \)[/tex] of a binomial distribution is given by:
[tex]\[ \mu = n \cdot p \][/tex]
Substituting the values:
[tex]\[ \mu = 25 \cdot 0.4 = 10.0 \][/tex]
#### Variance
The variance [tex]\( \sigma^2 \)[/tex] of a binomial distribution is given by:
[tex]\[ \sigma^2 = n \cdot p \cdot (1 - p) \][/tex]
Substituting the values:
[tex]\[ \sigma^2 = 25 \cdot 0.4 \cdot (1 - 0.4) = 25 \cdot 0.4 \cdot 0.6 = 6.0 \][/tex]
#### Standard Deviation
The standard deviation [tex]\( \sigma \)[/tex] is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \][/tex]
Substituting the variance:
[tex]\[ \sigma = \sqrt{6.0} \approx 2.449 \][/tex]
So, the mean is [tex]\( 10.0 \)[/tex], the variance is [tex]\( 6.0 \)[/tex], and the standard deviation is approximately [tex]\( 2.449 \)[/tex].
### Part ii: Probability Calculations
#### (a) [tex]\( P(X \geq 20) \)[/tex]
To find [tex]\( P(X \geq 20) \)[/tex], we use the cumulative distribution function (CDF) of the binomial distribution. The CDF up to 19 gives us [tex]\( P(X \leq 19) \)[/tex], so:
[tex]\[ P(X \geq 20) = 1 - P(X \leq 19) \][/tex]
From the computations, [tex]\( P(X \leq 19) \approx 0.999946410253622 \)[/tex], so:
[tex]\[ P(X \geq 20) = 1 - 0.999946410253622 \approx 5.359 \times 10^{-5} \][/tex]
#### (b) [tex]\( P(X \leq 5) \)[/tex]
The probability [tex]\( P(X \leq 5) \)[/tex] can be found directly using the CDF of the binomial distribution:
[tex]\[ P(X \leq 5) \approx 0.02936 \][/tex]
#### (c) [tex]\( P(X = 10) \)[/tex]
The probability [tex]\( P(X = 10) \)[/tex] is given by the probability mass function (PMF) of the binomial distribution:
[tex]\[ P(X = 10) \approx 0.16116 \][/tex]
### Summary:
- Mean: [tex]\( 10.0 \)[/tex]
- Variance: [tex]\( 6.0 \)[/tex]
- Standard Deviation: [tex]\( 2.449 \)[/tex]
- [tex]\( P(X \geq 20) \approx 5.359 \times 10^{-5} \)[/tex]
- [tex]\( P(X \leq 5) \approx 0.02936 \)[/tex]
- [tex]\( P(X = 10) \approx 0.16116 \)[/tex]
### Part i: Mean, Variance, and Standard Deviation of [tex]\( X \)[/tex]
Given:
- Probability of a worker not having health insurance, [tex]\( p = 0.4 \)[/tex].
- Number of workers in the sample, [tex]\( n = 25 \)[/tex].
#### Mean
The mean [tex]\( \mu \)[/tex] of a binomial distribution is given by:
[tex]\[ \mu = n \cdot p \][/tex]
Substituting the values:
[tex]\[ \mu = 25 \cdot 0.4 = 10.0 \][/tex]
#### Variance
The variance [tex]\( \sigma^2 \)[/tex] of a binomial distribution is given by:
[tex]\[ \sigma^2 = n \cdot p \cdot (1 - p) \][/tex]
Substituting the values:
[tex]\[ \sigma^2 = 25 \cdot 0.4 \cdot (1 - 0.4) = 25 \cdot 0.4 \cdot 0.6 = 6.0 \][/tex]
#### Standard Deviation
The standard deviation [tex]\( \sigma \)[/tex] is the square root of the variance:
[tex]\[ \sigma = \sqrt{\sigma^2} \][/tex]
Substituting the variance:
[tex]\[ \sigma = \sqrt{6.0} \approx 2.449 \][/tex]
So, the mean is [tex]\( 10.0 \)[/tex], the variance is [tex]\( 6.0 \)[/tex], and the standard deviation is approximately [tex]\( 2.449 \)[/tex].
### Part ii: Probability Calculations
#### (a) [tex]\( P(X \geq 20) \)[/tex]
To find [tex]\( P(X \geq 20) \)[/tex], we use the cumulative distribution function (CDF) of the binomial distribution. The CDF up to 19 gives us [tex]\( P(X \leq 19) \)[/tex], so:
[tex]\[ P(X \geq 20) = 1 - P(X \leq 19) \][/tex]
From the computations, [tex]\( P(X \leq 19) \approx 0.999946410253622 \)[/tex], so:
[tex]\[ P(X \geq 20) = 1 - 0.999946410253622 \approx 5.359 \times 10^{-5} \][/tex]
#### (b) [tex]\( P(X \leq 5) \)[/tex]
The probability [tex]\( P(X \leq 5) \)[/tex] can be found directly using the CDF of the binomial distribution:
[tex]\[ P(X \leq 5) \approx 0.02936 \][/tex]
#### (c) [tex]\( P(X = 10) \)[/tex]
The probability [tex]\( P(X = 10) \)[/tex] is given by the probability mass function (PMF) of the binomial distribution:
[tex]\[ P(X = 10) \approx 0.16116 \][/tex]
### Summary:
- Mean: [tex]\( 10.0 \)[/tex]
- Variance: [tex]\( 6.0 \)[/tex]
- Standard Deviation: [tex]\( 2.449 \)[/tex]
- [tex]\( P(X \geq 20) \approx 5.359 \times 10^{-5} \)[/tex]
- [tex]\( P(X \leq 5) \approx 0.02936 \)[/tex]
- [tex]\( P(X = 10) \approx 0.16116 \)[/tex]
