Question

Estimate the indicated probability by using the normal approximation as an approximation to the binomial distribution. (PROBLEMS 4 & 5) 4. Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that of 10,000 randomly selected hair dryers, at least 219 are defective.

5. In one county, the conviction rate for speeding is 85%. Estimate the probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

Answer 1

4

`P(X \ge 219 ) = 0.093`

5

`P(X \ge 219 ) = 0.1038`

Explanation:

From the question we are told that

The percentage of hair dryers that are defective is p=2%
`= 0.02`

The sample size is
`n =10000`

The random number is x = 219

The mean of this data set is evaluated as

`\mu = n* p`

substituting values

`\mu = 10000 * 0.02`

`\mu =200`

The standard deviation is evaluated as

`\sigma = √([\mu (1 -p)])`

substituting values

`\sigma = √([200 (1 -0.02)])`

`\sigma = 14`

Since it is a one tail test the degree of freedom is

df = 0.5

So

`x_l = x- 0.5`

`x = 219 - 0.5`

`x = 218.05`

Now applying normal approximation

`P(X \ge 219 ) = P(z > (x - \mu)/(\sigma) )`

substituting values

`P(X \ge 219 ) = P(z > (218.5 - 200)/(14) )`

`P(X \ge 219 ) = P(z > 1.32} )`

From the z-table

`P(X \ge 219 ) = 0.093`

Considering Question 5

The random number is x = 90

The mean is
`\mu = n * p`

Where n = 100

and p = 0.85

So

`\mu = 0.85 *100`

`\mu = 85`

The standard deviation is evaluated as

`\sigma = √([\mu (1 -p)])`

substituting values

`\sigma = √([85 (1 -0.85)])`

`\sigma = 3.5707`

Since it is a one tail test the degree of freedom is

df = 0.5

Now applying normal approximation

`P(X \ge 90 ) = P(z > (x - \mu)/(\sigma) )`

substituting values

`P(X \ge 219 ) = P(z > (89.5 - 85)/(3.5707) )`

`P(X \ge 219 ) = P(z > 1.2603} )`

From the z-table

`P(X \ge 219 ) = 0.1038`