Question 1081885: According to recent surveys, 60% of households have personal computers. If a random sample of 188 households is selected, what is the probability that more than 69 but fewer than 98 have a personal computer? Round z value calculations to 2 decimal places and final answer to 4 decimal places.
P(69 < X < 98)
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! 60% of households have computers. So, of 188 60% potentially have computers. Did you calculate 60% of 188? You can do it in your head.
First, 60% is 60 out of every hundred, that's what percent means, comes from per centum. So if you want to calculate 60% of 1, or 12, or any other number, first you take 60 per-centum and make it 60 per-one. If it's 60 for 100, you divide by 100 and get 0.6 per one. Now you can apply it to any number. 188 x 0.6 = 18.8 x 6.
Now, 8 x 6 = 48, write down 8 and we carry 4. Next 8 x 6 = 48 plus 4 = 52, write 2 and carry 5. last 6 x 1 = 6 plus 5 we carry = 11 We've got 1128. We have one decimal, makes our number 112.8. Now you decide if 112.8 is more than 69 but fewer than 98.
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