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

Concepts and reason

The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean and any particular normal distribution is completely specified by the numbers is mean and standard deviation.

Fundamentals

The formula of normal probability is, $Z = \frac{{\bar x - \mu }}{\sigma }$

Here, $\mu {\rm{ and }}\sigma$ is normal distribution mean and standard deviation.

a)

The mean value is less than the median value, so based on this evidence to say that the distribution of scores on this final exam is left skewed.

b)

The half of the data lies above median and half the data lies below median,

Hence, the above 70 score, because the median at 74 is at the 50^th percentile. (That is, half above, half below)

c)

Calculate the probability randomly chosen student scored above 75.

$\begin{array}{c}\\P\left( {X > 75} \right) = 1 - P\left( {Z \le 75} \right)\\\\ = 1 - P\left( {Z \le \frac{{75 - 70}}{{10}}} \right)\\\\ = 1 - P\left( {Z \le 0.5} \right)\\\\ = 1 - \left( {{\rm{ = NORMSDIST(0}}{\rm{.5)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel function}}} \right)\\\\ = 1 - 0.6915\\\\ = 0.3085\\\end{array}$

d)

Calculate the probability the average score for a random sample of 40 students is above 75.

$\begin{array}{c}\\P\left( {X > 75} \right) = 1 - P\left( {Z \le 75} \right)\\\\ = 1 - P\left( {Z \le \frac{{75 - 70}}{{10/\sqrt {40} }}} \right)\\\\ = 1 - P\left( {Z \le 3.16} \right)\\\\ = 1 - \left( {{\rm{ = NORMSDIST(3}}{\rm{.16)}}} \right){\rm{ }}\left( {{\rm{Use MS Excel function}}} \right)\\\\ = 1 - 0.9992\\\\ = 0.0008\\\end{array}$

e)

Calculate the cutting the sample size in half act the standard error of the mean.

$\begin{array}{c}\\{\rm{Lower}} = \frac{\sigma }{{\sqrt n }}\\\\ = \frac{{10}}{{\sqrt {40} }}\\\\ = 1.581\\\end{array}$

$\begin{array}{c}\\{\rm{Upper}} = \frac{\sigma }{{\sqrt n }}\\\\ = \frac{{10}}{{\sqrt {20} }}\\\\ = 2.236\\\end{array}$

Ans: Part a

The distribution of scores on this final exam is left skewed.

Part b

Since half of the data lies above median and median > 70 so we expect most students to score above 70 points.

Part c

The probability randomly chosen student scored above 75 is, 0.3085.

Part c

The probability the average score for a random sample of 40 students is above 75 is, 0.0008.

Part e

The cutting the sample size in half act the standard error of the mean is, $\left( {1.581{\rm{ to }}2.236} \right).$