Question

suppose 56% of the students in a university are baseball players. if a sample of 840 students is selected, what is the probability that the sample proportion of baseball players will be greater than 51% ? round your answer to four decimal places.

Answer 1

The probability that the sample proportion of baseball players will be greater than 51%, to find this we can use the normal approximation to the binomial distribution. The probability is approximately 0.6594 (rounded to four decimal places).

Step-by-step explanation:

To calculate the probability that the sample proportion of baseball players will be greater than 51%, we can use the normal approximation to the binomial distribution. First, we need to find the mean and standard deviation of the sample proportion.

Given that 56% of the students are baseball players and the sample size is 840, we can calculate:

Mean: (56/100) * 840 = 470.4

Standard Deviation: sqrt((56/100) * (44/100) * 840) = 12.07

Next, we need to standardize the desired sample proportion of 51% using the z-score formula:

Z-Score: (0.51 - 0.56) / 0.1207 = -0.4138

Finally, we can use a standard normal distribution table or a calculator to find the probability that the z-score is greater than -0.4138. The corresponding probability is approximately 0.6594. Therefore, the probability that the sample proportion of baseball players will be greater than 51% is 0.6594 (rounded to four decimal places).