Answer:
The value is [tex]P(X \le 60 ) = 0.00135[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 90 \ minutes[/tex]
The standard deviation is [tex]\sigma = 10[/tex]
Generally 1 hours = 60 minutes
Generally the probability that the exam will be completed in one hour or less is
[tex]P(X \le 60 ) = 1 - P(X > 60 )[/tex]
Here
[tex]P(X > 60 ) = P( \frac{ X - \mu }{ \sigma } > \frac{ 60 - 90 }{ 10 } )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
[tex]P(X > 60 ) = P( Z> -3 )[/tex]
From the z -distribution table the probability of ( Z> -3 )
[tex]P( Z> -3 ) = 0.99865[/tex]
[tex]P(X > 60 ) = 0.99865[/tex]
=> [tex]P(X \le 60 ) = 1 - 0.99865[/tex]
=> [tex]P(X \le 60 ) = 0.00135[/tex]
