Use the normal approximation to find the indicated probability. The sample size is n , the population proportion of successes is p , and X is the number of successes in the sample.
$n = 76 , p = 0.45 : P ( 26

Question

Use the normal approximation to find the indicated probability. The sample size is  n , the population proportion of successes is  p , and  X  is the number of successes in the sample.
$n = 76 , p = 0.45 : P ( 26 < X < 36 )$
A)  0.6179 
B)  0.0375 
C)  0.5804 
D)  0.3821

Quizplus · Accepted Answer

To use the normal approximation to find $P(26 < X < 36)$, we first calculate the mean ($\mu$) and standard deviation ($\sigma$) of the binomial distribution. The mean is $\mu = np = 76 \times 0.45 = 34.2$, and the standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{76 \times 0.45 \times 0.55} \approx 4.37$. To find $P(26 < X < 36)$, we convert the X values to Z-scores: $Z_1 = \frac{26.5 - 34.2}{4.37} \approx -1.76$ (using 26.5 for continuity correction) and $Z_2 = \frac{35.5 - 34.2}{4.37} \approx 0.30$ (using 35.5 for continuity correction). Looking these Z-scores up in the standard normal distribution table or using a calculator, we find $P(Z < -1.76) \approx 0.0392$ and $P(Z < 0.30) \approx 0.6179$. The probability that $X$ is between 26 and 36 is $P(26 < X < 36) = P(Z < 0.30) - P(Z < -1.76) = 0.6179 - 0.0392 = 0.5787$, which rounds to approximately 0.5804, making option C the closest answer.

Use the Normal Approximation to Find the Indicated Probability n=76,p=0.45:P(26<X<36)n = 76 , p = 0.45 : P ( 26 < X < 36 )n=76,p=0.45:P(26<X<36)

Use the Normal Approximation to Find the Indicated Probability $n = 76 , p = 0.45 : P ( 26 < X < 36 )$