Question

the census bureau reports that 82% of americans over the age of 25 are high school graduates. a survey of randomly selected residents of certain county included 1360 who were over the age of 25, and 1100 of them were high school graduates. (a) find the mean and standard deviation for the number of high school graduates in groups of 1360 americans over the age of 25. mean = 1115.2 standard deviation = 14.13 (b) is that county result of 1100 unusually high, or low, or rather? (enter high or low or neither) neither note: you can earn partial credit on this problem.

Explanation:

Step1: Calculate the mean

The formula for the mean of a binomial distribution is $\mu = np$. Here, $n = 1360$ (number of trials) and $p=0.82$ (probability of success).
$\mu=1360\times0.82 = 1115.2$

Step2: Calculate the standard deviation

The formula for the standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$.
Substitute $n = 1360$ and $p = 0.82$ into the formula:
$\sigma=\sqrt{1360\times0.82\times(1 - 0.82)}=\sqrt{1360\times0.82\times0.18}=\sqrt{200.736}\approx14.17$ (Note: there might be a rounding difference in the given answer of 14.13, but using exact calculation steps as above).

Step3: Determine if 1100 is unusual

We use the range rule of thumb. Values are considered unusual if they are outside of $\mu\pm2\sigma$.
$\mu - 2\sigma=1115.2-2\times14.17 = 1115.2 - 28.34=1086.86$
$\mu + 2\sigma=1115.2 + 2\times14.17=1115.2+28.34 = 1143.54$
Since $1086.86<1100<1143.54$, 1100 is within the range of $\mu\pm2\sigma$.

Answer:

(a) Mean = 1115.2, Standard deviation $\approx14.17$ (b) neither