Question

If xis a binomial random variable, compute P(x for each of the following cases:(c) P(x<2),n=9,p=0.2

P(x<2)=?

Answer 1

Explanation:

P(x < 2) = P(x = 0) + P(x = 1)

= (0.8)⁹ + 9(0.8)⁸(0.2)

≈ 0.4362.

Answer 2

P(x < 2) = 0.4362 (4 d.p.)

Explanation:

To compute P(x < 2) for a binomial random variable with parameters n = 9 and p = 0.2, use the cumulative distribution function (CDF) of the binomial distribution.

P(x < 2) means the probability that x takes on a value less than 2. This can be calculated as:

`P(x < 2) = P(x = 0) + P(x = 1)`

The binomial probability formula is:

`\boxed{\displaystyle P(x = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^(n - k)}`

Calculate P(x = 0):

`\displaystyle P(x = 0) = \binom{9}{0} \cdot 0.2^0 \cdot (1 - 0.2)^(9 - 0)`

`\displaystyle P(x = 0) = (9!)/(0!(9-0)!) \cdot1 \cdot (0.8)^(9)`

`\displaystyle P(x = 0) = 1 \cdot1 \cdot 0.134217728`

`\displaystyle P(x = 0) = 0.134217728`

Calculate P(x = 1):

`\displaystyle P(x = 1) = \binom{9}{1} \cdot 0.2^1 \cdot (1 - 0.2)^(9 - 1)`

`\displaystyle P(x = 1) = (9!)/(1!(9-1)!) \cdot 0.2 \cdot (0.8)^(8)`

`\displaystyle P(x = 1) = 9 \cdot 0.2 \cdot 0.16777216`

`\displaystyle P(x = 1) = 0.301989888`

Now, add these probabilities together:

`P(x < 2) = P(x = 0) + P(x = 1)`

`P(x < 2)= 0.134217728 + 0.301989888`

`P(x < 2)= 0.436207616`

`P(x < 2)=0.4362\; \sf (4\;d.p.)`

So, P(x < 2) for the given binomial random variable with n = 9 and p = 0.2 is approximately 0.4362 (rounded to four decimal places).