Week 3 Assignment: Apply Probability, Sampling Distributions, and Inference

Probability and sampling distribution
Name
Institutional affiliations
Chapter Five
Quiz 1
Solution.
Mean=2500
Standard deviation=500
By using the 68-95-99.7 rule, nearly 68% of the light bulbs have a life time between:
Mean- standard deviation=2500-500
=2000
Mean+ standard deviation=2500+500
=3000
An approximate number of 95% have a life of between
Mean-2 standard deviations =2500-(2*500)
= 1500 and
Mean+2 standard deviations=2500+ (2*500)
=3500
95.7% of the bulbs have a life of between
Mean-3 standard deviations=2500-(3*500)
=1000 and
Mean +3 standard deviations=2500+3 standard deviations
=2500+ (3*500)
=4000
This means that 49.85% (95.7%/2) of the bulbs have a life that is inferior to 2500 (mean)
Z-score = value – (Mean / SD)
=2500-2500
500
=0
From the z table 50% is the corresponding percentile of bulbs with life of less than
2500 hours.
Quiz 2
Solution
Mean = 370
Standard deviation=5,
Standard deviation = 1
68-95-99.7 rule states that:
68% of all the observations lie within 1 SD of the mean.
=370-5
=365 hours AND
=370+5
=375 hours.
Z (for 365 value) = (365-370)/5
=-1 which corresponds to 15.87% percentile of the Z score table
Z (for the 375 value) = (375-370)/5
=1 which corresponds to 84.13% percentile of the Z score table
Therefore:
84.13%-15.87% = 68.26% of all scores fall between 1 standard deviation of the mean on either side.
Question 3
Solution:
Mean=$60
Standard deviation= $12
60-12 = 48 and 60+12 = 72
$48 and $72
Quiz 4
Solution
Mean = $50, Standard deviation = $10
Data value = $50/$10
25th percentile z-score =-0.

674
-0.674= (y – 50) / 10
6.74 = (y – 50)
y=50 – 6.74
y= $43.26
Jen’s monthly phone bills are less than $43.26
Quiz 5
Solution
Mean = 0.30 Standard deviation=0.01
Use the 68-95-99.7 rule.
=0.30 + 0.01
=0.31(68% above the mean)
=0.30 + 2(0.01)
=0.32(95% above the mean)
Area above the mean with 2 standard deviations=0.9772.
Percentage of bolts with D>0.32=1-0.9772.
=0.0228 equivalent to 2.28%
Quiz 6
Solution
Mean =88, Standard deviation= 10, Sample size=25
Z score= (90.8-88)/10
=0.28
0.28 corresponds to 11.03% or an overall 61.03% of the population above the mean by use of a Z score table.
For a random sample of 25 years, 25*0.6103=15 years will rainfall be less than 90.8 inches.
Quiz 7
Solution
Z score = (70 – 73) / 1.58
= – 1.899
Area of the Z score=0 .02872.87% of the sample means are below than 70.
Quiz 8
Solution
Solution:
Mean = 50, SD = 10
Use the 68-95-99.7 rule.
a)
Mean –standard deviation=50-10=40 and mean+ standard deviation=50+10=60
40 -60 fall between -1 and +1 standard deviation.
b)
68%
c)
50-2(10) =30 and 50+2(10) =70
30-70
d)
13.5%

Chapter Six
Quiz 1
Solution:
a)
A 12 can only occur when all the dice roll number six.
Number of outcomes for each dice=6
Probability (6) = 1 / 6.
Probability of 2 sixes when both dice are rolled:
(1 / 6) * (1 / 6) = 1 / 36 = 0.0277.
Since chances of 0.0277 <= 0.05 the probability of rolling a 12 is significant using 2 dice.
b)
Probability of late arrival= P (late) / P (on time or late)
This translates to 20 / 500 = 0.04. Since 0.04 <= 0.05 a bus arriving late arriving is significant
Quiz 2
Solution
Number of outcomes =8
Number of outcomes without tail=1
Probability of not getting tail=1/7
Probability of getting at least one head i.e. 1 or more heads= 7/8
=0.875 which is equivalent to 87.5%
Quiz 3
Solution
For equally likely events, the probability of each is 1/64 = 0.0156 equivalent to 1.6%

Quiz 4
Solution
Total marbles=4+3+7
=14 marbles
Probability of a blue marble picked randomly =3/14
=0.2243 OR 21.4%

Quiz 5
Solution
Sample size=20
Number of occurrences for 98,000 =14
Probability of selection=14/20
=0.7
Quiz 6
Probability of head= 1 in 4 throws
In 32 throws= 32*1/4
=32/4
=8
Quiz 7
Solutions
a)
The expectancy of life for a female with an age of 60 is an average of 23.21 years or 83
Years.
The expectancy of life for a female with an age of 70 is an average of 12.98 years or 82
Years.
b)
The probability of dying is 0.00774 for a female with an age of 60 in the next year.This means an average of 90,847 * 0.00774 = 703 may die during that year.
Number of female leaving at the age of 61= 90,847 – 703 = 90,144.
The probability of a male of the age of 70 to die in the next year is 0.0289.
This means that an average of 70,214 * 0.0289 = 2,029 would die during that year.
Total of male living at the age of 71= 70,214 – 2,029
= 68,185.