Answer:
0.2805
Step-by-step explanation:
Given that p = 20% = 0.2, n = 300.
The mean (μ) = np = 0.2 * 300 = 60
The standard deviation (σ) = [tex]\sqrt{np(1-p)}=\sqrt{300*0.2(1-0.2)}=6.93[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma }[/tex]
For x = 57:
[tex]z=\frac{57-60}{6.93 } =-0.43[/tex]
For x = 62:
[tex]z=\frac{62-60}{6.93 } =0.29[/tex]
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
