Question

A magazine reported that 6% of American drivers read the newspaper while driving. If 300  drivers are selected at random, find the probability that exactly 25 say they read the Newspaper while driving.

Answer 1

Here

p = 0.06

q = 0.94 and

n = 300

Step1: Check whether a normal approximation can be used

$np=300\times 0.06=18\\nq=300\times 0,94=292$

Since $np\geq5\ \ and \ \ nq\geq5$ , we can use the normal distribution.

Step2: To find the mean and standard deviation,

$Mean\ \ \mu=np=18\\Standard\ Deviation\ \ \sigma=\sqrt{npq}=\sqrt{300(0.06)(0.94)}=4.11$

Step3: Write the probability notation. $P(X=25)$

Step4: Rewrite using the continuity correction factor.

$P(24.5<X<25.5)$

Step5: Find the corresponding z values.

$z=\dfrac{24.5-18}{4.11}=1.58\ \ \ ,z=\dfrac{25.5-18}{4.11}=1.82$

Step6: Find the solution.

The area between the two z value is

$0.9656-0.9429 = 0.0227, or \ 2.27 \%$

Hence, the probability that exactly 25 people read the newspaper while driving is 2.27%