STATISTICS: Statistical analyses with TI-83 – Part II:  Distributions

 

 

Problem 1:

Given a discrete probability distribution f(x) of a r.v. X

1. Find the expected value  E(X) of X

2. Find the standard deviation s of  X

 

Example:  Distribution of  X is in the table below. Find m = E(X) and  s = sd(X)

 

x	f(x)
0	0.018
1	0.218
2	0.473
3	0.182
4	0.109

Solution:

1.       Press STAT, then 1 (Edit); ENTER.

2.       Type x - values in L1 and probabilities  f(x) in L2. Press ENTER after every entry. 

3.       Press STAT, choose CALC and press 1 (1-Var Stats); ENTER

4.       Press 2nd  then 1 (L1) 

5.       Press , (comma)  and then 2nd and 2 (L1) 

6.       Press ENTER

Display:

·         The expected value is marked (wrongly)  x-bar and equals  m = E(X) = 2.146

·         The standard deviation is marked  s_x  and equals  s = sd(X) = 0.942

 

 

 

Problem 2:   Binomial probabilities

 

If X is binomial with n trials ac probability pf the success  p, then

• binompdf(n,p,x)  calculates  P(X = x)
• binomcdf(n,p,x)  calculates  P(X ≤ x)

  

Example:  Given that X has a binomial distribution with  n = 25 and p = 0.17 , find

A.     P(X ≤ 5)

B.     P(X = 5)

Solution:  A

1.       Press 2nd  and then VARS (DISTR)

2.       Select DISTR and then  A: binomcdf(    ENTER.

3.       Type  25,.17,5)   ENTER

4.       The answer is  0.75753

Solution:  B

1.       Press 2nd  and then VARS (DISTR)

2.       Select DISTR and then  0: binompdf(    ENTER.

3.       Type  25,.17,5)   ENTER

4.       The answer is  0.18161

 

 

 

 

 

 

 

 

Problem 3:  Normal probabilities and percentiles

 

If X is normal with mean  m and standard deviation s, then 

• normalcdf(a,b,m,s)  calculates  P(a ≤ X ≤ b)

- to calculate P(X ≤ b) let  a =  -10⁹⁹  

     (that is    (-) 1 2nd  , (EE)  99  which displays   -1E99)

- to calculate P(X ≥ a) let  b =  10⁹⁹  (that is   1 2nd  , (EE)  99 )

 

• invNorm(p,m,s)  calculates  the 100p-th percentile of X, that is the value c such that P(X ≤ c) = p

 

Example:  Given that X has a normal distribution with  m= 7.5 and s = 2.1, find

A.     P(3 < X ≤ 11)

B.     P(X ≤ 5)

C.     P( X ≥ 9)

D.     The 95^th percentile, that is the value  c such that  P(X<c) = 0.95

 

Solution:  A

1.       Press 2nd  and then VARS (DISTR)

2.       Select DISTR and then  choose  2: normalcdf(    ENTER.

3.       Type  3,11,7.5,2.1)   ENTER

4.       The answer is  0.93615

Solution:  B

1.       Press 2nd  and then VARS (DISTR)

2.       Select DISTR and then  2: normalcdf(    ENTER.

3.       Type  -1E99,5,7.5,2.1)   ENTER

4.       The answer is  0.11693

Solution:  C

1.       Press 2nd  and then VARS (DISTR)

2.       Select DISTR and then  2: normalcdf(    ENTER.

3.       Type  9, 1E99,7.5,2.1)   ENTER

4.       The answer is  0.23753

 

Solution:  D

5.       Press 2nd  and then VARS (DISTR)

6.       Select DISTR and then  3: invNorm(    ENTER.

7.       Type  .95,7.5,2.1)   ENTER

8.       The answer is  10.95

 

 

 

Reference: TI-83 Plus, Texas Instrument Inc., 1999