Question

The probability of rain on the last day of July is 95 % . If the probability remains constant for the first seven days of August , what is the probability that it will rain at most two of those seven days in August ? Find the Mean , Standard Deviation , and Variance .

Answer 1

The probability that it rains at most 2 days is 0.00005995233 and the variance is 0.516

The probability that it rains at most 2 days

The given parameters are:

• Number of days, n = 7
• Probability that it rains, p = 95%
• Number of days it rains, x = 2 (at most)

The probability that it rains at most 2 days is represented as:

P(x ≤ 2) = P(0) + P(1) + P(2)

Each probability is calculated as:

$P(x) = ^nC_x * p^x * (1 - p)^{n - x}$

So, we have:

$P(0) = ^7C_0 * (92\%)^0 * (1 - 92\%)^{7 - 0} = 0.00000002097$

$P(1) = ^7C_1 * (92\%)^1 * (1 - 92\%)^{7 - 1} = 0.00000168821$

$P(2) = ^7C_2 * (92\%)^2 * (1 - 92\%)^{7 - 2} = 0.00005824315$

So, we have:

P(x ≤ 2) =0.00000002097 + 0.00000168821 + 0.00005824315

P(x ≤ 2) = 0.00005995233

Hence, the probability that it rains at most 2 days is 0.00005995233

The mean

This is calculated as:

Mean = np

So, we have:

Mean = 7 * 92%

Evaluate

Mean = 6.44

Hence, the mean is 6.44

The standard deviation

This is calculated as:

σ = √np(1 - p)

So, we have:

σ = √7 * 92%(1 - 92%)

Evaluate

σ = 0.718

Hence, the standard deviation is 0.718

The variance

We have:

σ = 0.718

Square both sides

σ² = 0.718²

Evaluate

σ² = 0.516

This represents the variance

Hence, the variance is 0.516

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