Assume that x is a binomial random variable with n = 1000 and p = 0.80. Use a normal approximation to find $\mathrm { P } ( 800

Question

Assume that x is a binomial random variable with n = 1000 and p = 0.80. Use a normal approximation to find $\mathrm { P } ( 800 < \mathrm { x } \leq 830 )$
A)0.4920
B)0.4741
C)0.0753
D)0.4760

Quizplus · Accepted Answer

Using the normal approximation, the mean is $ \mu = np = 800 $ and the standard deviation is $ \sigma = \sqrt{np(1-p)} = \sqrt{160} \approx 12.65 $. Applying continuity correction, we find $ P(800.5 < x \leq 830.5) $. Converting to the standard normal distribution, we calculate $ P\left(\frac{800.5-800}{12.65} < Z \leq \frac{830.5-800}{12.65}\right) $, which gives $ P(0.04 < Z \leq 2.41) $. Using standard normal distribution tables or a calculator, this probability is approximately 0.4760.

Assume That X Is a Binomial Random Variable with N P(800<x≤830)\mathrm { P } ( 800 < \mathrm { x } \leq 830 )P(800<x≤830)

Assume That X Is a Binomial Random Variable with N $\mathrm { P } ( 800 < \mathrm { x } \leq 830 )$