Expert: Ahmed Salami - 10/23/2004

Question
This topic is about sampling distribution. The question is as follow:

A study shoes that 46 percent of all employees of TGB Corporation are less willing to give up free time today than they were 5 years ago. What's the probability that in a random sample of 25 employees, more than 50 percent will say that they are less willing to give up free time today than they were 5 years ago?

My own working is as follow but I do not understand and know if its right.

Let x be the random variables that they are less willing to go.

P= 100%-50% = 50% with n = 25
np= 25 x 0.5 = 17.5 > 5

Where sigma p = (p(1-p)/n)^1/2 = (0.01) ^1/2 = 0.1

0.5 - 0.1 = 0.4

From Z table P(ps > 0.4) = 0.3446

The answer given is correct but I do not understand the working. Your lucid explanation will be very much appreciated.

Answer
Hi Jason,
There were a lot of mistakes in your calculations(it doesn't matter if you still got the answer right).
Firstly, np = 25 x 0.46 = 11.5 which is still > 5
sigma = [p.(1-p).n]^1/2  [and not (p(1-p)/n)^1/2]
     = [(0.46)(0.54)(25)]^1/2
     = 2.49
the score for 50% = 0.5 x 25 = 12.5
Therefore,
z = (12.5 - 11.5)/2.49
 = 0.4
From z tables,
P = 0.1554 for z = 0.4
The probability for z > 0.4 is therefore 0.5 - 0.1554
= 0.3446
I hope it is clear.
Regards.