02 0215 -3 -2-1 01 2 3 For a certain law school, the entering students have an a

Question

02 0215 -3 -2-1 01 2 3 For a certain law school, the entering students have an average LSAT score of about with a standard deviation of about 40. The smoothed density histogram of the scores appears to follow a normal distribution. 700 LSAT 1. Sketch a smoothed density histogram (a bell-shaped curve) of the distributions for the LSAT scores. Don't spend too much time on this question. Just sketch the curve with 700 as the mean. It should look similar to the curve in the upper right hand comer of this page. 780 220 620 660 700 740 780 MeaN Because this distribution is approximately normal, the mean is equal to the median. Therefore, the mean will also divide our data into two equal pieces. We will use the mean as our measure of center. Locate the mean on the distribution and mark it.- 00 3. What fraction of the data is to the left of the mean? How much is to the right of the mean? The frachoN to the left and right is the same so the 4. Calculate the z (standardized) score for an LSAT score of 700. Write that value here. (You may know this without any calculations) Remember that a z-score tells you how many standard deviations away from the mean a value is. If the value is the mean what does the z-score have to be? 5. Find the interval of LSAT scores that is 1 standard deviation away from the mean. Write the interval here and mark it on your smoothed histogram. 700 +40 and 700-40

Explanation / Answer

Given that,

Population mean (mu) = 700

Population standard deviation (sigma) = 40

18) Here we have to find P(Z > 0.5).

This we can find in excel.

Syntax :

=1 – NORMSDIST(z)

Where z is z-score

P(Z > 0.5) = 0.3085 = 30.85%

19) The number of students scored higher than Joe = 9000*0.3085 = 2776.84

20) Here we have to find x.

X we can find by using formula :

X = mu + z*sigma

X = 700 + (-1.6)*40 = 636

Now we have to find P(X < 636) i.e. we have to find P(Z < -1.6)

P(Z < -1.6) = 0.0548 = 0.0548*100 = 5.48%

22) Here also we have to find x when probability = 15% = 0.15

Here we have to find z for 0.15

This also we can find in excel.

Syntax :

=NORMSINV(probability)

Z = -1.04

X = 700 + (-1.04) * 40 = 658.54

23) Number of students will be required to attend this program are 0.0548*9000 = 493.19

24) Here also we have to find x when probability = 1 – 0.01 = 0.99

We use same technique as 21).

Z = 2.33

X = 700 + (2.33) * 40 = 793.05

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.