Answer:
4
[tex]P(X \ge 219 ) = 0.093[/tex]
5
[tex]P(X \ge 219 ) = 0.1038[/tex]
Step-by-step explanation:
From the question we are told that
The percentage of hair dryers that are defective is p=2% [tex]= 0.02[/tex]
The sample size is [tex]n =10000[/tex]
The random number is x = 219
The mean of this data set is evaluated as
[tex]\mu = n* p[/tex]
substituting values
[tex]\mu = 10000 * 0.02[/tex]
[tex]\mu =200[/tex]
The standard deviation is evaluated as
[tex]\sigma = \sqrt{[\mu (1 -p)]}[/tex]
substituting values
[tex]\sigma = \sqrt{[200 (1 -0.02)]}[/tex]
[tex]\sigma = 14[/tex]
Since it is a one tail test the degree of freedom is
df = 0.5
So
[tex]x_l = x- 0.5[/tex]
[tex]x = 219 - 0.5[/tex]
[tex]x = 218.05[/tex]
Now applying normal approximation
[tex]P(X \ge 219 ) = P(z > \frac{x - \mu}{\sigma} )[/tex]
substituting values
[tex]P(X \ge 219 ) = P(z > \frac{218.5 - 200}{14} )[/tex]
[tex]P(X \ge 219 ) = P(z > 1.32} )[/tex]
From the z-table
[tex]P(X \ge 219 ) = 0.093[/tex]
Considering Question 5
The random number is x = 90
The mean is [tex]\mu = n * p[/tex]
Where n = 100
and p = 0.85
So
[tex]\mu = 0.85 *100[/tex]
[tex]\mu = 85[/tex]
The standard deviation is evaluated as
[tex]\sigma = \sqrt{[\mu (1 -p)]}[/tex]
substituting values
[tex]\sigma = \sqrt{[85 (1 -0.85)]}[/tex]
[tex]\sigma = 3.5707[/tex]
Since it is a one tail test the degree of freedom is
df = 0.5
Now applying normal approximation
[tex]P(X \ge 90 ) = P(z > \frac{x - \mu}{\sigma} )[/tex]
substituting values
[tex]P(X \ge 219 ) = P(z > \frac{89.5 - 85}{3.5707} )[/tex]
[tex]P(X \ge 219 ) = P(z > 1.2603} )[/tex]
From the z-table
[tex]P(X \ge 219 ) = 0.1038[/tex]
