Question 1: Use the standard normal distribution table to find the indicated areas under the standard normal curve:

a. Area under the curve between z=0 and z = 2.15
b. Area under the curve between z=0 and z = -1.55
c. Area under the curve to the right of z = 0.48
d. Area under the curve to the left of z = -.78
e. Area under the curve between z=0.93 and z = 3.21

Solution: (a) We need to compute the following probability:

\[\Pr \left( 0\le Z\le 2.15 \right)=\Pr \left( Z\le 2.15 \right)-\Pr \left( Z\le 0 \right)\]

\[={0.9842}-{0.5}={0.4842}\]

where this probability is computed used the NORMSDIST procedure from Excel.

(b) Now, we need to compute the following probability:

\[\Pr \left( -1.55\le Z\le 0 \right)=\Pr \left( Z\le 0 \right)-\Pr \left( Z\le -1.55 \right)\]

\[={0.5}-{0.0606}={0.4394}\]

(c) Now, we need to compute the following probability:

\[\Pr \left( Z\ge {0.48} \right)=1-\Pr \left( Z\le 0.48 \right)=1-{0.6844}={0.3156}\]

(d) Finally, we need to compute the following probability:

\[\Pr \left( Z\le {-0.78} \right)={0.2177}\]

Question 2: The price of shares of Bank of Florida at the end of trading each day for the last year followed the normal distribution. Assume there were 240 trading days in the year. The mean price was $42.00 per share and the standard deviation was $2.25 per share. (Round your answers to 2 decimal places. Omit the "$" and "%" signs in your response.)
(a1) What percent of the days was the price over $45.00?
(a2) How many days would you estimate?
(b) What percent of the days was the price between $38.00 and $40.00?
(c) What was the stock’s price on the highest 15 percent of days?

Solution: (a1) We need to compute the following probability:

\[\Pr \left( X\ge {45} \right) = \Pr \left( \frac{X-{42}}{2.25}\ge \frac{{45}-{42}}{2.25} \right) = \Pr \left( Z\ge 1.3333 \right) = 1-\Pr \left( Z\le 1.3333 \right) = 1-{0.9088} = {0.0912}\]

which corresponds to approximately 9.12%.

(a2) The expected number is 0.0912*240 = 21.888 \(\approx 22\) days.

(b) We need to compute the following probability:

\[\Pr \left( {38}\le X\le {40} \right) = \Pr \left( \frac{{38}-{42}}{2.25}\le \frac{X-{42}}{2.25}\le \frac{{40}-{42}}{2.25} \right)\] \[= \Pr \left( -1.7778\le Z\le -0.8889 \right) = \Pr \left( Z\le -0.8889 \right)-\Pr \left( Z\le -1.7778 \right) = {0.187}-{0.0377} = {0.1493}\]

which corresponds to 14.93%.

(c) Finally, we need to compute the following:

\[U=42+{{z}_{0.15}}\times 2.25=42+1.0364\times 2.25=44.332\]

which corresponds to approximately $44.33.

Question 3: A student is enrolled in an introductory programming class and a communications class at the university. If the student in the programming class takes a midterm exam and earns a score of 76, while the student in communications class takes a midterm exam and earns a score of 72. In the programming class, the class mean was 64 and the standard deviation was 8. In the communications class, the class mean was 60 with a standard deviation of 7.5.

In which class did the student perform better compared to the rest of the students in the class? Show your work to support your decision. Assume test scores are normally distributed.

(Hint: How many standard deviations are the student’s midterm scores away from the respective class means?)

a. Programming Class Z-score

b. Communications Class Z-score

c. Explain which student did better and why

Solution: (a) The z-score for the programming class is z = (76 – 64)/8 = 1.5

(b) The z-score for the communication class is z = (72 - 60)/7.5 = 1.6.

(c) The student did better in the communications class, because the z-score is higher.

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