Question

1. a final exam for a college algebra class had 60 questions. for all the students that took the exam, the mean number of questions answered correctly is 54, with a standard deviation of 12. would it be reasonable to assume that the distribution of the number of questions answered correctly is approximately normal? explain.
2. the website random.org has a random decimal generator that will generate a list of random numbers between 0 and 1.

a. what would be the shape of the density curve for the distribution of random numbers generated? draw a picture.
b. what proportion of the random numbers generated would be between 0.6 and 0.8?

Explanation:

Step1: Recall normal - distribution properties

For a normal distribution, most of the data should be within a reasonable range around the mean. The range of values for a normal distribution is often considered to be within 3 standard - deviations of the mean.

Step2: Calculate the range within 3 standard - deviations

The lower limit is $\mu - 3\sigma$ and the upper limit is $\mu+3\sigma$, where $\mu = 54$ and $\sigma = 12$.
Lower limit: $54-3\times12=54 - 36=18$.
Upper limit: $54 + 3\times12=54+36 = 90$.
But the number of questions is between 0 and 60. Since the upper - limit of the 3 - standard - deviation range (90) is outside the possible range of scores (0 - 60), it is not reasonable to assume the distribution is approximately normal.

Step3: Analyze the uniform distribution for question 4a

The random decimal generator on random.org generates numbers between 0 and 1. The density curve for a uniform distribution on the interval $[a,b]=[0,1]$ is a horizontal line. The height of the density curve for a uniform distribution $U(a,b)$ is given by $f(x)=\frac{1}{b - a}$. Here, $a = 0$ and $b = 1$, so $f(x)=1$ for $0\leq x\leq1$ and $f(x)=0$ otherwise. The shape of the density curve is a horizontal line at $y = 1$ over the interval $[0,1]$.

Step4: Calculate the proportion for question 4b

For a uniform distribution $U(a,b)$ on the interval $[a,b]$, the probability $P(c\leq X\leq d)$ is given by $\frac{d - c}{b - a}$, where $a = 0$, $b = 1$, $c = 0.6$, and $d = 0.8$.
$P(0.6\leq X\leq0.8)=\frac{0.8 - 0.6}{1-0}=0.2$.

Answer:

1. It is not reasonable to assume the distribution is approximately normal because the upper - limit of the 3 - standard - deviation range (90) is outside the possible range of scores (0 - 60).

4a. The shape of the density curve is a horizontal line at $y = 1$ over the interval $[0,1]$.
4b. 0.2