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The top web browser in 2015 was Chrome with 51.74% of the market. In a random sample of 250 people, what is the probability that fewer than 110 did not use Chrome?
Intro Stats / AP Statistics
Chapter 6
The Normal Distribution
Section 4
The Normal Approximation to the Binomial Distribution
Probability Topics
Missouri State University
Piedmont College
Boston College
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Okay, Number 18. We have an equals to 50. Okay. And in the problem, it said 51.74% used chrome. However, the probability they're asking is for the ones that did not use chrome. Okay, so we have to take the compliment. So it's one minus 0.5174 Okay, I converted 51.74% 2 decimal, and we get point for 8 to 6. Okay, so that's RPI value. All right, So for the mean, it's just n times P 2 50 times, 500.48 to 6, and we get 1 20.65 Okay. And for the standard deviation, it's a square root of NP, which we said was 1 20.65 And that's gonna be times one minus p, which we actually have already. It's 0.5174 But anyhow, for the standard deviation, you should get 7.90 rounded to two decimals. Okay, We want to find the probability that X is less than 1 10 Okay? And we have to look at our conversion short. When X is less than a number, you subtract 0.5, so This is equivalent to X less than 109.5. Okay, subtract 0.5 from 1 10 And that gives you the new X value. Now for the Z score, we have X minus the mean that whole thing divided by the standard deviation. And you should get negative 1.41 Okay, you look up that Z score on the Z score table. Negative. 1.41 and it will give you a value of point 07 93 Okay, Now, this is for when x is less than 109.5. And that's exactly what we want. So converting this to decimal move the desk from place to place is and you should get seven 0.93% and that's number 18.
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