Question

Let X be a binomially distributed random variable with parameters n=500 and p=0.3. The probability that X is no larger than one standard deviation above its mean is closest to which of the following?

a. 0.579
b. 0.869
c. 0.847
d. 0.680

Answer 1

c. 0.847

Explanation:

From the given information;

`X \sim Binomial (500,0.3)`

Given that n = 500 which is too large, binomial distribution can now be approximated to

`N ( \mu , \sigma^2)`

where;

`\mu = np`

`\mu =500 * 0.3`

`\mu =150`

`\sigma^2= np(1-p)`

`\sigma^2= 500 * 0.3(1-0.3)`

`\sigma^2= 150(0.7)`

`\sigma^2= 105\\`

∴

`P(X \leq \mu + \sigma ) = ( (X-\mu)/(\sigma) \leq 1)`

`P(X \leq \mu + \sigma ) = ( Z \leq 1)`

`P(X \leq \mu + \sigma ) =\Phi (1)`

From the z table

`P(X \leq \mu + \sigma ) =0.841`

Thus, our value is closest to the option c which 0.847