Answer :
Answer:
The value is [tex]P(X \ge 15) = 0.5[/tex]
Step-by-step explanation:
From the question we are told that
The probability of the device failing during the warranty period is [tex]p = 0.005[/tex]
The sample size is [tex]n = 3000[/tex]
The random variable considered is x = 15
Generally this is distribution is binomial given the fact that there is only two out comes hence
X which is a variable representing a randomly selected selected electronic follows a binomial distribution i.e
[tex]X \~ \ B(n , p)[/tex]
Now the mean is mathematically evaluated as
[tex]\mu = n * p[/tex]
=> [tex]\mu = 3000 * 0.005[/tex]
=> [tex]\mu =15[/tex]
The standard deviation is mathematically represented as
[tex]\sigma = \sqrt{np(1 -p )}[/tex]
=> [tex]\sigma = \sqrt{3000 * 0.005 * (1 - 0.005 )}[/tex]
=> [tex]\sigma = 3.86[/tex]
Now given that n is very large, then it mean that we can successfully apply normal approximation on this binomial distribution
So
[tex]P(X \ge 15) = P( \frac{X - \mu}{\sigma } \ge \frac{x - \mu}{\sigma } )[/tex]
Now applying Continuity Correction we have
[tex]P(X \ge (15-0.5)) = P( \frac{X - \mu}{\sigma } > \frac{(15 -0.5) - 15}{3.86 } )[/tex]
Generally [tex]\frac{X - \mu}{\sigma } = Z (The \ standardized \ value \ of \ X)[/tex]
[tex]P(X \ge (15-0.5)) = P(Z >-0.130 )[/tex]
From the z-table
[tex]P(Z >-0.130 ) =0.5[/tex]
Thus
[tex]P(X \ge 15) = 0.5[/tex]