Download Given a standardized normal distribution with a mean of 0 and a

Given a standardized normal distribution with a mean of 0 and a standard deviation of 1, what is the
probability that Z is less than 1.57?
.9418
.0582
.5000
1
Answer:
P(Z<1.57) = 0.9418 (From table of areas under the standard normal curve)
Question 2 3 points Save
The breaking strength of plastic bags used for packaging produce is normally distributed, with a
mean of 5 pounds per square inch and a standard deviation of 1.5 pounds per square inch. What
proportion of the bags have a breaking strength between 5 and 6.5 pounds?
.5
.34
.75
.24
Answer: Let X denote the breaking strength of a randomly selected plastic bag. Then X follows the
normal distribution with mean 5 and standard deviation 1.5. Let Z = (X-5)/1.5. The Z follows the
standard normal distribution.
When X = 5, Z = (5-5)/1.5 = 0/1.5 =0
When X = 6.5, Z = (6.5-5)1.5 = 1.5/1.5 =1
P(5<X<6.5)=P(0<Z<1)=P(Z<1)-P(Z<0)=0.8413-0.5=0.3413
The proportion bags with breaking strength between 5 pounds and 6.5 pounds is 0.3413.
It is 0.34 correct to 2 decimal places.
Question 3 3 points Save
A DBU campus program evenly enrolls undergraduate and graduate students. If a random sample of
4 students is selected from the program to be interviewed about the introduction of a
Chick-fil-A food outlet on the ground floor of the Collins building, what is the probability that all 4
students selected are undergraduate students?
0.0256
0.0625
0.16
1.00
Answer:
P(a randomly selected student is undergraduate)=0.5
P(all 4 students are undergraduate)=P(first student is undergraduate)* P(second student is
undergraduate)* P(third student is undergraduate)* P( student is undergraduate)=0.5 * 0.5 * 0.5 *
0.5 = 0.0625
Question 4 3 points Save
True or False: The number of customers arriving at a department store in a 5-minute period has a
binomial distribution.
True
False
Answrer: False
The number of customers arriving at a department store in a 5-minute period has a Poisson
distribution.
Question 5 3 points Save
Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A or B) =
Answer:
P(A or B) = P(A)+P(B)-P(A and B)=0.4+0.5-0.1=0.8
Question 6 3 points Save
There are 47 contestants at a national dog show. How many different ways can contestants fill the
first place, second place, and third place positions?
Answer: 47*46*45 = 97290
The first place can be filled by any one of the 47 contestants.
The second place can be filled by any one of the remaining 46 contestants.
The third place can be filled by any one of the remaining 45 contestants.
All the three places can be filled by 47*46*45=97290 ways.
Question 7 3 points Save
A high school debate team of 4 is to be chosen from a class of 35. How many possible ways can the
team be formed?
Answer:
Number of combinations of 35 objects taken 4 at a time = Combin(35,4)= 35!/(4!(35-4)!)
= (35*34*33*32)/(1*2*3*4) = 52360
The team can be formed in 52360 different ways.
Question 8 3 points Save
Given that X is a normally distributed random variable with a mean of 50 and a standard deviation of
2, find the probability that X is between 47 and 54.
Answer:
X follows normal distribution with mean 50 and standard deviation 2.
Z = (X-50)/2 follows the standard normal distribution.
When X=47, Z=(47-50)/2 = -3/2 = -1.5
When X=54, Z=(54-50)/2=4/2=2
P(47<X<54)=P(-1.5<Z<2)=P(Z<2)-P(Z<-1.5)=0.9772-0.0668=0.9104
Question 9 3 points Save
If there are 6 balls in a hat and 5 are blue and 1 is red. What are the chances of drawing the red ball
4 times out of 4 draws if you are sampling with replacement?
Answer:
Since we are sampling with replacement , the outcomes of the successive draws are independent.
The probability of drawing red ball in any draw is 1/6
P(red ball in all 4 draws) = P(red ball in first draw) * P(red ball in second draw) * P(red ball in third
draw) * P(red ball in fourth draw) = (1/6)* (1/6)* (1/6)* (1/6)=1/(6*6*6*6) =1/1296=0.000772
Question 10 3 points Save
Assume you are randomly guessing on a multiple choice test with answers A B C and D. Assume
there are 5 question on the test. What is the chance that you will miss all 5 questions?
Answer:
P(miss all 5 questions)=P(miss first question) * P(miss second question) * P(miss third question) *
P(miss fourth question)* P(miss fifth question) = (3/4)*(3/4)*(3/4)*(3/4)*(3/4)
=(3*3*3*3*3)/(4*4*4*4*4) = 243/1024 = 0.2373
Question 11 3 points Save
Assume you are randomly guessing on a multiple choice test with answers A B C and D. Assume
there are 5 question on the test. What is the chance that you will get all 5 questions correct?
Answer:
P(get all 5 questions)=P(get first question) * P(get second question) * P(get third question) * P(get
fourth question)* P(get fifth question) = (1/4)*(1/4)*(1/4)*(1/4)*(1/4)
=1/(4*4*4*4*4) = 1/1024 = 0.000977
Question 12 3 points Save
Assume you are randomly guessing on a multiple choice test with answers A B C and D. Assume
there are 5 question on the test. What is the chance that you will get 4 out of 5 questions correct?
Answer:
P(4 out of 5 questions correct) = P(first question wrong and remaining questions correct)
+ P(second question wrong and remaining questions correct)
+ P(third question wrong and remaining questions correct)
+ P(fourth question wrong and remaining questions correct)
+ P(fifth question wrong and remaining questions correct)
=(3/4)*(1/4)*(1/4)*(1/4)*(1/4) + (3/4)*(1/4)*(1/4)*(1/4)*(1/4) +(3/4)*(1/4)*(1/4)*(1/4)*(1/4)
+ (3/4)*(1/4)*(1/4)*(1/4)*(1/4) + (3/4)*(1/4)*(1/4)*(1/4)*(1/4)
=3/1024 +3/1024 +3/1024 +3/1024 +3/1024 = 15/1024 = 0.0146
Question 13 3 points Save
The pH level in a shampoo
Continuous
Discrete
Answer: Continuous
Question 14 3 points Save
Rolling a single die 34 times, keeping track of the numbers that are rolled.
Not binomial: the trials are not independent
Not binomial: there are too many trials
Procedure results in a binomial distribution
Not binomial: there are more than two outcomes for each trial
Answer: Not binomial: there are more than two outcomes for each trial
Question 15 3 points Save
The mean is µ = 15.2 and the standard deviation is s = 0.9. Find the probability that X is greater than
16.1.
0.8413
0.1587
0.1550
0.1357
Answer:
Z = (X-15.2)/0.9
When X = 16.1, Z=(16.1-15.2)0.9 = 0.9/0.9 =1
P(X>16.1)=P(Z>1) =1-P(Z<1)=1- 0.8413=0.1587
Question 16 3 points Save
The mean is µ = 22.0 and the standard deviation is s = 2.4. Find the probability that X is between 19.7
and 25.3.
0.3370
0.4107
1.0847
0.7477
Answer:
Z=(X-22.0)/2.4
When X = 19.7, Z=(19.7-22.0)/2.4 = -0.9583
When X=25.3, Z=(25.3-22.0)/2.4=1.375
P(19.7<X<25.3)=P(-0.9583<Z<1.375)=P(Z<1.375)-P(Z<-0.9583)=0.9154-0.1689=0.7465
Question 17 3 points Save
The mean is µ = 15.2 and the standard deviation is s = 0.9. Find the probability that X is greater than
17.
0.9713
0.9821
0.9772
0.0228
Answer:
Z=(X-15.2)/0.9
When X=17, Z=(17-15.2)/0.9=1.8/0.9=2
P(X>17)=P(Z>2)=1-P(Z<2)=1-0.9772=0.0228
Question 18 3 points Save
Use the confidence level and sample data to find a confidence interval for estimating the population
µ. A random sample of 94 light bulbs had a mean life of X = 587 hours with a standard deviation of s
= 36 hours. Construct a 90 percent confidence interval for the mean life, µ, of all light bulbs of this
type.
Answer:
Confidence coefficient = 1-α=0.90
Α=1-0.90=0.10=10%
10% critical value of t with 93 degrees of freedom = 1.6614
Lower confidence limit = 587 – 1.6614 * 36/√94 = 587 – 6.17 = 580.83
Upper confidence limit = 587 + 1.6614 * 36/√94 = 587 + 6.17 = 593.17
The 90% confidence interval is (580.93 hours, 593.17 hours)
Question 39 3 points Save
Assume that women have heights that are normally distributed with a mean of 63.6 and a standard
deviation of 2.5 inches. Find the value of the third quartile.
64.3 inches
66.1 inches
65.3 inches
67.8 inches
Answer:
X = height of a randomly selected woman.
X follows normal distribution with mean 63.6 and standard deviation 2.5
Z = (X-63.6)/2.5 follows standard normal distribution.
Third quartile of Z is 0.6745
Third quartile of X is 63.6 + 2.5 * 0.6745 = 65.2862 = 65.3 inches(correct to 1 decimal place)
Question 40 3 points Save
A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To
test this claim against the alternative that the actual proportion of doctors who recommend aspirin
is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend
aspirin. The value of the test statistic in this problem is approximately equal to:
- 4.12
- 2.33
- 1.86
- 0.07
Sample proportion = p^ = 83/100 =0.83
Population proportion under the null hypothesis = 9/10 = 0.90
Sample size n = 100
The test statistic is Z = (p^ - p0)/√(p0*(1-p0)/n) = (83/100 – 9/10)/√((9/10)*(1-9/10)/100)
=(0.83-0.90)/√(0.90*0.10/100) = -0.07/√(0.09/100) = -0.07/√(0.0009) = -0.07/0.03 =-2.3333
=-2.33(correct to 2 decimal places)
Question 41 3 points Save
Determine if the following is a probability distribution.
Yes
No
Data not given
Question 42 3 points Save
Use the given degree of confidence and sample data to construct a confidence interval population
proportion p. n = 87, x = 48; 98 percent
0.448 30
H0: µ = 30
H1: µ 30
H1: µ = 30
Answer: 2% critical value of Z is 2.3263.
Sample proportion = p^ = x/n = 48/87 = 0.5517
Lower confidence limit = 0.5517 – 2.3263 * √(0.5517*(1-0.5517)/87) = 0.5517-0.1240 =0.4277
Upper confidence limit = 0.5517 + 2.3263 * √(0.5517*(1-0.5517)/87) = 0.5517+0.1240 =0.6758
The confidence interval is (0.4277, 0.6758)
Question 47 3 points Save
If z is a standard normal variable, find P(-0.73
Answer: The question is not complete.