Question

Problem #5: Nine percent of men and 0.25% of women cannot distinguish between the colors red and green. This is the type of color blindness that causes problems with traffic signals. (a) If 9 men are randomly selected for a study of traffic signal perceptions, find the probability that between 2 and 4 inclusive of them have this type of color blindness. (b) In a group of 180 men, find the mean number that are color blind. (c) In a group of 180 men, find the standard deviation of the number that are color blind. (d) Suppose that a group of 180 men are randomly selected, and 27 of them are color blind. Is this a significantly high number that would perhaps suggest that the given percentage of men that are color blind (i.e., 9%) is not correct?

problem 6 :

The percentage of adults who have at some point in their life been told that they have hypertension is 23.53%. In a sample of 14 adults, let X be the number who have been told that they have hypertension. Consider the following probability distribution for X. x P(x) 0 ? 1 ? 2 ? 3 ? 4 ? 5 a 6 0.0596 7 0.0210 8 0.0056 9 0.0012 10 0.0002 11 0.0000 12 0.0000 13 0.0000 14 0.0000

(a)	Find the missing entry that is labelled as 'a'.

(b)

Suppose that a group of 14 adults are randomly selected, and 6 of them have been told that they have hypertension. Is this a significantly high number that would suggest that the given percentage of adults who have been told that they have hypertension (i.e., 23.53%) is not correct? (A) Yes, because 0.0596 is greater than .05. (B) Yes, because 0.1291 is greater than .05. (C) No, because 6 a not a lot more than expected. (D) Yes, because 6 a lot more than expected. (E) Yes, because 0.0876 is greater than .05. (F) No, because 0.1291 is greater than .05. (G) No, because 0.0876 is greater than .05. (H) No, because 0.0596 is greater than .05.

Answer 1

5)

n=9

p=0.0025

a)

P ( X = 2) = C (9,2) * 0.09^2 * ( 1 - 0.09)^7=      0.1507
P ( X = 3) = C (9,3) * 0.09^3 * ( 1 - 0.09)^6=      0.0348
P ( X = 4) = C (9,4) * 0.09^4 * ( 1 - 0.09)^5=      0.0052

P(2≤x≤4) = P(X=2) + P(X=3) + P(X=4) =0.1906

b)

Mean = np =    180   *   0.090   =           16.200

c)

Standard deviation = √(np(1-p)) =   √   14.7420   =               3.8395

d)

The range rule of thumb suggests that values are unusual if they lie outside of these limits:                              
( µ ± 2σ) = (   16.2   ± 2*   3.8395   ) = (   8.5209   ,   23.88 )

27 lies outside this range

so, answer:  Yes, because 27 a lot more than expected.

6)

n=14

p=0.2353

Binomial probability is given by
P(X=x) = C(n,x)*px*(1-p)(n-x)

X	P(X)
0	0.0234
1	0.1007
2	0.2015
3	0.2480
4	0.2098
5	0.1291
6	0.0596
7	0.0210
8	0.0056
9	0.0012
10	0.0002
11	0.0000
12	0.0000
13	0.0000
14	0.0000

value of a = 0.1291

b)

Mean = np =    14   *   0.235   =           3.294
Standard deviation = √(np(1-p)) =   √   2.5191   =               1.5872

The range rule of thumb suggests that values are unusual if they lie outside of these limits:                              
( µ ± 2σ) = (   3.2942   ± 2*   1.5872   ) = (   0.1199   ,   6.469 )

answer:  No, because 6 a not a lot more than expected.