Answer:
![\mu_p -\sigma_p = 0.74-0.0219=0.718](https://tex.z-dn.net/?f=%20%5Cmu_p%20-%5Csigma_p%20%3D%200.74-0.0219%3D0.718)
![\mu_p +\sigma_p = 0.74+0.0219=0.762](https://tex.z-dn.net/?f=%20%5Cmu_p%20%2B%5Csigma_p%20%3D%200.74%2B0.0219%3D0.762)
68% of the rates are expected to be betwen 0.718 and 0.762
![\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696](https://tex.z-dn.net/?f=%20%5Cmu_p%20-2%2A%5Csigma_p%20%3D%200.74-2%2A0.0219%3D0.696)
![\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784](https://tex.z-dn.net/?f=%20%5Cmu_p%20%2B2%2A%5Csigma_p%20%3D%200.74%2B2%2A0.0219%3D0.784)
95% of the rates are expected to be betwen 0.696 and 0.784
![\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674](https://tex.z-dn.net/?f=%20%5Cmu_p%20-3%2A%5Csigma_p%20%3D%200.74-3%2A0.0219%3D0.674)
![\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806](https://tex.z-dn.net/?f=%20%5Cmu_p%20%2B3%2A%5Csigma_p%20%3D%200.74%2B3%2A0.0219%3D0.806)
99.7% of the rates are expected to be betwen 0.674 and 0.806
Step-by-step explanation:
Check for conditions
For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:
a) Independence : we assume that the random sample of 400 students each student is independent from the other.
b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.
c) np= 400*0.74= 296>10
n(1-p) = 400*(1-0.74)=104>10
So then we have all the conditions satisfied.
Solution to the problem
For this case we know that the distribution for the population proportion is given by:
![p \sim N(p, \sqrt{\frac{p(1-p)}{n}})](https://tex.z-dn.net/?f=%20p%20%5Csim%20N%28p%2C%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%29)
So then:
![\mu_p = 0.74](https://tex.z-dn.net/?f=%20%5Cmu_p%20%3D%200.74)
![\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219](https://tex.z-dn.net/?f=%20%5Csigma_p%20%3D%5Csqrt%7B%5Cfrac%7B0.74%281-0.74%29%7D%7B400%7D%7D%3D0.0219)
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
![\mu_p -\sigma_p = 0.74-0.0219=0.718](https://tex.z-dn.net/?f=%20%5Cmu_p%20-%5Csigma_p%20%3D%200.74-0.0219%3D0.718)
![\mu_p +\sigma_p = 0.74+0.0219=0.762](https://tex.z-dn.net/?f=%20%5Cmu_p%20%2B%5Csigma_p%20%3D%200.74%2B0.0219%3D0.762)
68% of the rates are expected to be betwen 0.718 and 0.762
![\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696](https://tex.z-dn.net/?f=%20%5Cmu_p%20-2%2A%5Csigma_p%20%3D%200.74-2%2A0.0219%3D0.696)
![\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784](https://tex.z-dn.net/?f=%20%5Cmu_p%20%2B2%2A%5Csigma_p%20%3D%200.74%2B2%2A0.0219%3D0.784)
95% of the rates are expected to be betwen 0.696 and 0.784
![\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674](https://tex.z-dn.net/?f=%20%5Cmu_p%20-3%2A%5Csigma_p%20%3D%200.74-3%2A0.0219%3D0.674)
![\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806](https://tex.z-dn.net/?f=%20%5Cmu_p%20%2B3%2A%5Csigma_p%20%3D%200.74%2B3%2A0.0219%3D0.806)
99.7% of the rates are expected to be betwen 0.674 and 0.806