I need help with some elementary stats ASAP!

Question

The readings on the scientific thermometers are normally distributed with a mean of 0º C and a standard deviation of 1º C. This is an example of the standard normal distribution. Use StatCrunch or TI83/84 to find the following probabilities.

a)    Find the probability that, at the freezing point of water, the reading is less than 1.8º C.

b. Find the probability that, at the freezing point of water, the reading is above 1.8º C.

c. Find the probability that, at the freezing point of water, the reading is between 0º C and 2.00 º C.

d. Find the temperature at the 70^th  percentile: P₇₀, that is the temperature that separates the bottom 70% from the top 30%.

2. MCAT scores have an approximate normal distribution with mean at 500 and standard deviation 10.6. Use StatCrunch or TI83/84 to find the following probabilities.

a. What percent of students get scores over 510?

b. What percent of students get scores under 480?

c. What percent of students get scores between 480 and 520?

d. What is the Z-score of a MCAT score of 530? Based on the Z-score, is this an unusual score? Explain your answer.

e. If 9 test takers are randomly selected, what is the mean for the sample mean distribution?

f. If 9 test takers are randomly selected, what is the standard error for the sample mean distribution?

g. If 9 test takers, find the probability that the group has a MCAT score average above 507.

3. In this experiment, we will learn about the central limit theorem. You will do four sample distributions simulation using the APPLET function in StatCrunch. Open StatCrunch and click on the tab APPLETS. Down near the end of the list, choose “Sampling distributions”. On this window, choose “right-skewed” and change the upper bound in the window below to 10. Click COMPUTE (at the bottom) which brings up a new window. On this window change the sample size to 1 and press the “1000 times” button at the top. Fill in the information for the sample means distribution (which is the bottom tier on the window).

a. Mean: _______   Standard deviation: _________ Shape___________

In part b, change the sample size to 2 and repeat the procedure:

b. Mean: _______    Standard deviation: _________ Shape__________

In part c, change the sample size to 10 and repeat the procedure:

c. Mean: _______    Standard deviation: _________ Shape___________

In part b, change the sample size to 20 and repeat the procedure:

d. Mean: ________ Standard deviation: _________ Shape__________

e. What happens to the mean as the sample size increases from 1 to 2 to 10 to 20?

f. What happens to the standard deviation as the sample size increases from 1 to 2 to 10 to 20?

g. What happens to the distribution shape as the sample size increases from 1 to 2 to 10 to 20?

h. How do these results illustrate the Central Limit Theorem?         

Thank you so much in advance!!

Answer 1

The responses here are based on the TI -8X series graphing calculators with statistical functions. Normal distribution data is available thru StatCrunch at the following link: https://www.statcrunch.com/app/calculator.html?calculator=normal

Assumptions for question 1. Water is pure, air pressure is standard sea level.

Question 1

a. nmcdf(-1e3,-1.8,0,1) = 0.03593

b. nmcdf(1.8,1e3 ,0,1) = 0.03593

c. nmcdf(0,2,1,0) = 0.47255

d. invnm(.7,0,1) = 0.52440

Question 2.

a. nmcdf(510,1e3,500,10.6) = 0.1727

b. nmcdf(-1e3,480,500,10.6) = 0.02959

c. nmcdf(480,520,500,10.6) = 0.94081

d. Z = (530-500)/10.6. = 2.83. is it unusual? depends on your deffinition. If you define unusual as greater then 2 or less than -2, it is unusual. If you use 3 and -3 as your limits, it is not. Why is it unusual, because it represents a limit where the probability of the |Z| being greater than 2.83 is .00233

e. Answer - unknown, because it must be based on the SAMPLE taken. , although the expected value (at confidence level of 98%) should be 500 ± 24.659. [ note: the question probably was looking for 500 which is the mean of the expected sample mean distribution regardless of sample size, not the mean of a random sample of 9 unknown samples]

f. Answer - standard deviation of the mean (standard error) with a population sigma of 10.6 and a sample size of 9 = 10.6/3. or 3.5333

g. Answer - from an unknown group of 9 test takers, the probability of the actual mean value of the 9 tests being larger than 507 is nmcdf(507,1e3,500,3.5333) = 0.0238

Question 3 (using StatCrunch at https://www.statcrunch.com/app/

This is basically a lab exercise. Just follow the instructions with StatCrunch. Only can be done with StatCrunch as it depends on StatCrunch default values to create the skewed distribution. I do not have StatCrunch but the student is invited to follow the simple lab instructions.