Answer:
0.2805
Explanation:
Given that p = 20% = 0.2, n = 300.
The mean (μ) = np = 0.2 * 300 = 60
The standard deviation (σ) =
![√(np(1-p))=√(300*0.2(1-0.2))=6.93](https://img.qammunity.org/2022/formulas/mathematics/high-school/lx82ddjuidgt2t8bimiiiv48oodfyihmny.png)
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
![z=(x-\mu)/(\sigma )](https://img.qammunity.org/2022/formulas/mathematics/high-school/jchnxdqfozkmsf85nzuothzj4k49ijid1g.png)
For x = 57:
![z=(57-60)/(6.93 ) =-0.43](https://img.qammunity.org/2022/formulas/mathematics/high-school/pt6l4utt0er9iou8jczjuirpvwhxo1s1pa.png)
For x = 62:
![z=(62-60)/(6.93 ) =0.29](https://img.qammunity.org/2022/formulas/mathematics/high-school/expgplifmcivs27ym3ntf8b6ed0nkqc95v.png)
From the normal distribution table, P(57 < x < 62) = P(-0.43 < z < 0.29) = P(z < 0.29) - P(z < -0.43) = 0.6141 - 0.3336 = 0.2805