1.
Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children. Find the probability of exactly nine boys in ten births. Round the answer to the nearest thousandth.
A. |
D. 0.176 |
|
B. |
B. 0.0918 |
|
C. |
A. 0.800 |
|
D. |
C. 0.080 |
Find the area of the indicated region under the standard normal curve.
A. |
A. 0.3438 |
|
B. |
C. 1.3090 |
|
C. |
B. 0.6562 |
|
D. |
D. 0.3090 |
The cholesterol levels (in milligrams per deciliter) of 30 adults are listed below. Find the interquartile range for the cholesterol level of the 30 adults.
154 |
156 |
165 |
165 |
170 |
171 |
172 |
180 |
184 |
185 |
189 |
189 |
190 |
192 |
195 |
198 |
198 |
200 |
200 |
200 |
205 |
205 |
211 |
215 |
220 |
220 |
225 |
238 |
255 |
265 |
A. |
A. 211 |
|
B. |
C. 30 |
|
C. |
C. 31 |
|
D. |
B. 180 |
Given that P(A or B) = 1/5 , P(A) = 1/8, and P(A and B) = 1/9, find P(B).
A. |
77/360 B. |
|
B. |
67/360 C. |
|
C. |
11/180 D. |
|
D. |
A. 167/360 |
1: Success probability of boy p = 0.5
0.009
2: 0.6562
3: Interquartile Range = Third quartile - First Quartile
From data:
First Quartile = 180 and Third quartile = 211
Interquartile Range = 211 - 180
Interquartile Range = 31
4: Use this formula:
P(A or B) = P(A)+P(B) - P(A and B)
P(B) = P(A or B) + P(A and B) - P(A)
P(B) =1/5 + 1/9 - 1/8
P(B) = 67/360
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