Question

A company compiles data on a variety of issues in education. In 2004 the company reported that the national college​ freshman-to-sophomore retention rate was 66​%. Consider colleges with freshman classes of 500 students. Use the​ 68-95-99.7 rule to describe the sampling distribution model for the percentage of students expected to return for their sophomore years. Do you think the appropriate conditions are​ met

Answer 1

Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$  

And we have :

$\mu_p = 0.66$

$\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212$

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$  

And we have :

$\mu_p = 0.66$

$\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212$

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as 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)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).