Answer:
The probability is [tex]P(X \ge 20 ) = 0.3707[/tex]
Step-by-step explanation:
From the the question we are told that
The population proportion is p = 0.60
The sample size is n = 31
The mean is evaluated as
[tex]\mu = n * p[/tex]
substituting values
[tex]\mu = 31 *0.60[/tex]
[tex]\mu = 18.6[/tex]
The standard deviation is evaluated as
[tex]\sigma = \sqrt{n * p * (1- p )}[/tex]
substituting values
[tex]\sigma = \sqrt{31 * 0.6 * (1- 0.6 )}[/tex]
[tex]\sigma = 2.73[/tex]
The the probability that at least 20 of them have looked at their score in the past six months is mathematically represented as
[tex]P(X \ge 20) = 1- P(X < 20)[/tex]
applying normal approximation we have that
[tex]P(X \ge 20) = 1- P(X < (20-0.5))[/tex]
Standardizing
[tex]1 - P(X < 20) = 1 - P(\frac{X - \mu }{\sigma} < \frac{19.5 - \mu }{\sigma } )[/tex]
[tex]1 - P(X < 20) = 1 - P(Z < \frac{19.5 - 18.6 }{2.73 } )[/tex]
[tex]1 - P(X < 20) = 1 - P(Z < 0.33)[/tex]
Form the standardized normal distribution table we have that
[tex]P(Z < 0.0512)[/tex] = 0.6293
=> [tex]P(X \ge 20 ) = 1- 0.6293[/tex]
=> [tex]P(X \ge 20 ) = 0.3707[/tex]