Question

Each year about 1500 students take the introductory statistics course at a large university. This year scores on the nal exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the nal) but a few students scored below 20 points.
(a) Is the distribution of scores on this nal exam symmetric, right skewed, or left skewed?
(b) Would you expect most students to have scored above or below 70 points?
(c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution?
(d) What is the probability that the average score for a random sample of 40 students is above 75?
(e) How would cutting the sample size in half aect the standard error of the mean?

Answer 1

Answer:

a) Left-skewed

b) We should expect most students to have scored above 70.

c) The scores are skewed, so we cannot calculate any probability for a single student.

d) 0.08% probability that the average score for a random sample of 40 students is above 75

e) If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.

Step-by-step explanation:

To solve this question, we need to understand skewness,the normal probability distribution and the central limit theorem.

Skewness:

To undertand skewness, it is important to understand the concept of the median.

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

If the median is larger than the mean, the distribution is left-skewed.

If the mean is larger than the median, the distribution is right skewed.

Normal probabilty distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation, also called standard error of the mean $s = \frac{\sigma}{\sqrt{n}}$

(a) Is the distribution of scores on this nal exam symmetric, right skewed, or left skewed?

Mean = 70, median = 74. So the distribution is left-skewed.

(b) Would you expect most students to have scored above or below 70 points?

70 is below the median, which is 74.

50% score above the median, and 50% below. So 50% score above 74.

This means that we should expect most students to have scored above 70.

(c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution?

The scores are skewed, so we cannot calculate any probability for a single student.

(d) What is the probability that the average score for a random sample of 40 students is above 75?

Now we can apply the central limit theorem.

$\mu = 70, \sigma = 10, n = 40, s = \frac{10}{\sqrt{40}} = 1.58$

This probability is 1 subtracted by the pvalue of Z when X = 75. So

$Z = \frac{X - \mu}{\sigma}$

By the Central limit theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{75 - 70}{1.58}$

$Z = 3.16$

$Z = 3.16$ has a pvalue of 0.9992

1 - 0.9992 = 0.0008

0.08% probability that the average score for a random sample of 40 students is above 75

(e) How would cutting the sample size in half aect the standard error of the mean?

n = 40

$s = \frac{10}{\sqrt{40}} = 1.58$

n = 20

$s = \frac{10}{\sqrt{20}} = 2.24$

If the sample size is cut in half, the standard error of the mean would increase fro 1.58 to 2.24.