26 Jun I Need 100%
For this HW you should use Appendix Table for Normal Distribution
| Student | μ | n | p | Student | μ | n | p | |
| Acopan | 40 | 10 | 60% | Harris | 66 | 36 | 75% | |
| Ambol | 42 | 12 | 65% | Jacobs | 68 | 38 | 80% | |
| Anderson | 44 | 14 | 70% | Johnson | 70 | 40 | 85% | |
| Attai | 46 | 16 | 75% | Kurata | 72 | 42 | 90% | |
| Broseker | 48 | 18 | 80% | Mendoza | 74 | 44 | 95% | |
| Buchheit | 50 | 20 | 85% | Monyei | 76 | 46 | 82% | |
| Bullock | 52 | 22 | 90% | Munshi | 78 | 48 | 77% | |
| Caywood | 54 | 24 | 95% | Pickering | 65 | 25 | 72% | |
| Cox-Martinez | 56 | 26 | 82% | Rager | 80 | 50 | 60% | |
| Eickhoff | 58 | 28 | 77% | Ritchey | 82 | 48 | 65% | |
| Fox | 60 | 30 | 60% | Rosen | 84 | 44 | 70% | |
| Gray | 62 | 32 | 65% | Sekamwa | 86 | 42 | 75% | |
| Hampton | 64 | 34 | 70% | Turner | 88 | 36 | 80% |
Problem 1. For a Normal Distribution with the mean μ and a standard deviation σ = 25 calculate the z-scores that correspond to the following x-values: (Take value of from the table next to your name in the beginning of this homework). a) x = 65
b) x = 77 c) x = 82
Problem 2. Suppose x-value is normally distributed with a mean of and a standard deviation of 10. What is the probability that x will be less than 65?
(Take value of from the table above, next to your name).
Tip: Find z-value and then use Appendix Table for Normal Distribution
Problem 3. Assume the speed of vehicles on a certain road has an approximately
normal distribution with a mean mph and a standard deviation σ=10 mph. What is the probability that vehicles on this road will be going faster than 72 mph?
(Take value of from the table above, next to your name). Tip: Follow the same steps as in #2, but remember: P(z>a) = 1 – P(z<a)
Show work.
Problem 4. An average weight of rugby players can be described as Normal distribution with mean μ = 220 lbs. and standard deviation σ=20 lbs. If n randomly selected rugby players were selected (sample), what is the probability that the sample mean will be between 200 and 240 lbs.? (Take value of from the table next to your name in the beginning of this homework).
Tip: For the distribution of sample mean, standard deviation will be not 20. Apply formula for standard deviation of sample mean: σ/√n.
Show work.
Problem 5. Normal distribution has mean μ (see table in the beginning of this homework) and standard deviation σ=20.
Find the p% percentile for this distribution (use p assigned for you in the table posted in the beginning of this week homework).
Tip: From Appendix Table for Normal Distribution find z that has p% of area under the curve to the left. Then convert z to x, using formula: z = (x – )/σ.
Show work.
Binomial and Normal Distribution
Let’s practice with Appendix Tables for Binomial and Normal Distribution. Part 1. Table for Binomial Distribution applies to situation when each of our trials (attempts) has two possible results (outcomes): success or failure. We make a certain number of attempts (N trials) and each time we try, the probability of success is the same (p). For example, you flip the coin and count a number of times when you have a Head (success). In each flip (trial) p = 0.5. The Table gives us the probability that in n trials (attempts) we will succeed x times. All we need to find the probability is N (total number of trials), x (number of expected successes) and p (probability of success in each trial). For N=3, x=2 and p=0.10, probability is: 0.027.
In part 1, you should choose your own numbers for N and x. Keep probability of success p=0.6. Read details in the study material.
In Part 2 we practice with Appendix Table for Normal Distribution . The Table shows the calculated area under the Bell-Shaped curve to the left of a certain z-value. Z-value in the table goes from -3.4 to 3.4. Calculated Area gives us the probability that real z will be less than this z-value. For example, from the Table, calculated area to the left of z1=-2.95 is 0.0016
That means, P(z < z1) = P(z < -2.95) = 0.0016. For z2= – 2.87 we have: P(z < z2) = P(z < -2.87) = 0.0021 If you need P(z > -2.95), it will be 1 – 0.0016 = 0.9984 P(z > – 2.87) = 1 – 0.0021 = 0.9979 To find probability that z is between z1 and z2, use the formula: P(z1 < z < z2) = P(z < Z2) – p(z < Z1). For example, P(-2.95 < z < -2.87) = P(z < -2.87) – P(z < -2.95) = 0.0021 – 0.0016 = 0.0005 Read details in the study material.
In part 2, select your own values of z1 and z2, then use the Appendix Table for the Standard Normal Distribution to find : 1) P(z < z1) 2) P(z > z1) 3) P(z < z2) 4 )P(z > z2) 5) P(z1 < z < z2)
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