Question

Suppose 46% of the students in a university are baseball players. If a sample of 622 students is selected, what is the probability that the sample proportion of baseball players will be greater than 47%? Round your answer to four decimal places.

Answer 1

Answer:

Well by looking at this proportion the answer might be 1.0217

Step-by-step explanation:

46% of 622 = 286.12

47% of 622 = 292.34

292.34 / 286.12 = 0.0217...

Answer 2

Answer:

Probability of proportion will greater than 47% = 0.6915

Step-by-step explanation:

Given data:

population proportion p = 0.46

sample proportion b $\hat p = 0.47$

sample size be n = 622

Probability of proportion will greater than 47%

$P(\hat p > 0.47) = P[\fraca{\hat p - p}{\sqrt{p(1-p)/n}}]$

$= P[z > {\frac{0.47 -0.46}{\sqrt{0.46(1-0.46)/622}}]$

                          = P(z >0.500)

from standard z table

= 0.6915