Answer:
c. 0.847
Step-by-step explanation:
From the given information;
[tex]X \sim Binomial (500,0.3)[/tex]
Given that n = 500 which is too large, binomial distribution can now be approximated to
[tex]N ( \mu , \sigma^2)[/tex]
where;
[tex]\mu = np[/tex]
[tex]\mu =500 \times 0.3[/tex]
[tex]\mu =150[/tex]
[tex]\sigma^2= np(1-p)[/tex]
[tex]\sigma^2= 500 \times 0.3(1-0.3)[/tex]
[tex]\sigma^2= 150(0.7)[/tex]
[tex]\sigma^2= 105\\[/tex]
∴
[tex]P(X \leq \mu + \sigma ) = ( \dfrac{X-\mu}{\sigma} \leq 1)[/tex]
[tex]P(X \leq \mu + \sigma ) = ( Z \leq 1)[/tex]
[tex]P(X \leq \mu + \sigma ) =\Phi (1)[/tex]
From the z table
[tex]P(X \leq \mu + \sigma ) =0.841[/tex]
Thus, our value is closest to the option c which 0.847
