## Question 356:

1## Answer:

No answer provided yet.You'll need to find two areas under the normal curve given the mean of 151 and standard deviation 15.

- Find the normal deviate or z-score for the larger value : (155-151)/15 = a z-score of .266
- Find the area under the normal curve using the z-score to percentile calculator, choose 1-sided area. This provides an area of about 60.5%.
- Subtract the smaller area, which following the same steps above is (120-151)/15= -2.06 (notice the negative sign) and has area of ~1.96%. Subtracting the smaller gets us 60.5-1.96 = 58.54%

So we'd expect about 58.5% of the 500 students or 293 of them to have weights between 120 and 155 lbs. You can also visualize this area by taking a look at the interactive graph of the standard normal curve and enter in the mean of 151 and SD of 15 in the second calculator.