MATHEMATICS COLLEGE

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

Answers

Answer 1

Answer:

Answer:

c. 0.847

Step-by-step 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;

$\mu = np$

$\mu =500 * 0.3$

$\mu =150$

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

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

$\sigma^2= 150(0.7)$

$\sigma^2= 105\n$

∴

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

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

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

From the z table

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

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

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Answers

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