Common Probability Distributions

The standard deviation (σ) of a binomial distribution is calculated using the formula:

σ=Sqrt(n∗p∗(1−p)​)

Where:

• $\sigma$ (sigma) is the standard deviation.
• $n$ is the number of trials.
• $p$ is the probability of success.

In this case:

• $n$ (number of trials) = 100
• $p$ (probability of success) = 0.3

Now, calculate the standard deviation:

σ=sqrt(100∗0.3∗(1−0.3))​

σ=100∗0.3∗0.7​

σ=21​

$\sigma \approx 4.58$

So, the standard deviation of the binomial distribution with 100 trials and a probability of success of 0.3 is approximately 4.58.

Cumulative distribution function (CDF) of the exponential distribution is given by:

P(X≤x)=1−e(−λx)

Where:

• $P(X\le x)$ is the probability that the time between arrivals is less than or equal to $x$.
• $\lambda$ (lambda) is the rate parameter (0.2 arrivals per minute).
• $e$ is the base of the natural logarithm (approximately 2.71828).
• $x$ is the time interval (5 minutes in this case).

Now, plug in the values and calculate the probability:

P(X≤5)=1−e^(−0.2∗5)

P(X≤5)=1−e^(−1)≈1−0.36788≈0.6321

So, the probability that the time between two arrivals is less than 5 minutes is approximately 0.6321.

the formula for the z-score:

Z=(X−μ)/σ​

Where:

• Z is the z-score.
• X is the value you want to find the z-score for (in this case, 60).
• μ (mu) is the mean of the distribution (50).
• σ (sigma) is the standard deviation of the distribution (10).

Plug in the values:

Z=60−5010=1010=1Z=1060−50​=1010​=1

So, the z-score for a value of 60 in this normally distributed variable is 1.

Mean (μ) = n * p

In this case:

Number of trials (n) = 10 Probability of success (p) = 0.4

Mean (μ) = 10 * 0.4 = 4

So, the mean (expected value) of the binomial distribution is 4.