## STAT 13

(Sec. 1a-1c)

Introduction to Statistical Methods for the Life and Health Sciences

## Instructor: Ivo Dinov, Asst. Prof.

Departments of Statistics & Neurology

Due Date:

# Friday, Feb. 07, 2003, turn in after lecture

See the HW submission rules. On the front page include the following header.

(HW_4_1)
• The following table of Normal probabilities was obtained from Excel.
 x P(X≤x) x P(X ≤ x) x P(X ≤ x) 15 0.0000 20 0.0668 25 0.8413 16 0.0002 21 0.1587 26 0.9332 17 0.0013 22 0.3085 27 0.9772 18 0.0062 23 0.5000 28 0.9938 19 0.0228 24 0.6915 29 0.9987
Use the table to find the following when X~ Normal(μ = 23, σ2 = 4):
(i ) pr(X ≤19);  05. - 0.0228
(ii) pr(X < 19);  same
(iii) pr(X > 21);  1- 0.1587
(iv) pr(24 ≤ X ≤ 27) 0.9772 - 0.6915

• The following table of probabilities was obtained from STATA: Normal with mean μ= 8.6 and standard deviation σ= 1.28.
 P(X ≤ x) x 0.1000 6.9596 0.2500 7.7367 0.7500 9.4633 0.9000 10.2404
The number of liters of soft serve ice cream sold by an ice cream van driver in an afternoon is found to be Normally distributed with mean μ = 8.6 liters and standard deviation σ = 1.28 liters.

(i) What is the least amount of soft serve ice cream that is needed so that the driver can satisfy demand on 90% of afternoons?  Z=(X-8.6)/1.28,  Z (90%) = 1.28. Solve for X!

(ii) What is the interquartile range for the ice cream sales. (Central 50%), IQR = Q3-Q1,
Z1 = 0.675, Z2 = -0.675, Solve Z=(X-8.6)/1.28, for X1 and X2, IQR = X1 - X2.

• Use either STATA or a graphics calculator to solve the following problems where
X~Normal(μ = 5.1, σ2 = 0.872 ):
(i) What is the probability that X is greater than 6? Standardize: Z = (X-5.1)/0.87 = 0.9/0.87
P(Z > 0.9/0.87) = 0.5 - 0.3413
(ii) What is the probability that X is between 4.3 and 6.6? Z1 = (4.3 - 5.1)/0.87 = - 0.8/0.87
Z2 = (6.6 - 5.1)/0.87 = 1.5/0.87 ==> P(Z1 < Z < Z2) ~==  0.321 + 0.4573
(iii) What value of x gives P(X ≤x) = 0.45? Z = (X - 5.1)/0.87 = 1.645, Solve for X!
• X has a mean of -3 and a standard deviation of 5 and W has a mean of 5 and a standard deviation of 3. Let X and W be independent random variables and let Y = 3X - 3W.

•  (i) What are the mean and standard deviation of Y? E(Y) = 3 E(X) -3 E(W) = -9 -15 = -14

SD(Y) = Sqrt( 225 + 81) = 17.5
(ii) What can we say about the shape of the distribution of Y? If X and W are normally distributed, then so is Y!

• (HW_4_2) Suppose that X~ Normal(μ=3, σ2=16) compute the Z-scores for the following numbers and state how many SD's each of these numbers is away from μ:
• -5 ;  Z = (X-3)/4 = -2 (-2 SD's away from mean, below mean)
• 11;   Z = (11-3)/4 = 2
• 5;     Z = 1
• 1.4;   Z = -3/8
• What is the probability P(-3 ≤ X ≤ -1)? Is it different from P(-3 < X < -1)? Why? The two probabilities are the same, since the X ~ Normal Distribution and so X is continuous and including the limits does not change the chance. Z1 = (-3-3)/4 = -3/2,  Z2 = (-4)/4 = -1,  P(-3<X<-1) = 0.4332 - 0.3413.