Question

if ( x ) is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases
(a) ( n = 5, p = 0.8 )
( mu=)
( sigma^{2}=)
( sigma=)
(b) ( n = 5, p = 0.9 )
( mu=)
( sigma^{2}=)
( sigma=)
(c) ( n = 5, p = 0.1 )
( mu=)
( sigma^{2}=)
( sigma=)
(d) ( n = 5, p = 0.5 )
( mu=)
( sigma^{2}=)
( sigma=)

Explanation:

Step1: Recall the formulas for binomial distribution

For a binomial random variable \(X\) with parameters \(n\) (number of trials) and \(p\) (probability of success in a single trial):

• Mean \(\mu=np\)
• Variance \(\sigma^{2}=np(1 - p)\)
• Standard deviation \(\sigma=\sqrt{np(1 - p)}\)

Step2: Solve part (a) where \(n = 5\) and \(p=0.8\)

• Mean: \(\mu=np=5\times0.8 = 4\)
• Variance: \(\sigma^{2}=np(1 - p)=5\times0.8\times(1 - 0.8)=5\times0.8\times0.2 = 0.8\)
• Standard deviation: \(\sigma=\sqrt{0.8}\approx0.894\)

Step3: Solve part (b) where \(n = 5\) and \(p = 0.9\)

• Mean: \(\mu=np=5\times0.9=4.5\)
• Variance: \(\sigma^{2}=np(1 - p)=5\times0.9\times(1 - 0.9)=5\times0.9\times0.1 = 0.45\)
• Standard deviation: \(\sigma=\sqrt{0.45}\approx0.671\)

Step4: Solve part (c) where \(n = 5\) and \(p = 0.1\)

• Mean: \(\mu=np=5\times0.1 = 0.5\)
• Variance: \(\sigma^{2}=np(1 - p)=5\times0.1\times(1 - 0.1)=5\times0.1\times0.9=0.45\)
• Standard deviation: \(\sigma=\sqrt{0.45}\approx0.671\)

Step5: Solve part (d) where \(n = 5\) and \(p = 0.5\)

• Mean: \(\mu=np=5\times0.5=2.5\)
• Variance: \(\sigma^{2}=np(1 - p)=5\times0.5\times(1 - 0.5)=5\times0.5\times0.5 = 1.25\)
• Standard deviation: \(\sigma=\sqrt{1.25}\approx1.118\)

Answer:

(a) \(\mu = 4\), \(\sigma^{2}=0.8\), \(\sigma\approx0.894\)
(b) \(\mu = 4.5\), \(\sigma^{2}=0.45\), \(\sigma\approx0.671\)
(c) \(\mu = 0.5\), \(\sigma^{2}=0.45\), \(\sigma\approx0.671\)
(d) \(\mu = 2.5\), \(\sigma^{2}=1.25\), \(\sigma\approx1.118\)