Question 1

A slot machine at a casino pays out an average of $0.90,with a standard deviation of $120.It costs a dollar to play.If gamblers play this machine 6,969,600 times in a month,what is the probability that the casino will come out ahead? Assume that the casino's total profit follows a Normal model.
A)0.023
B)0.014
C)0.977
D)0.955
E)0.986

Accepted Answer

The answer of A slot machine at a casino pays...

Question 2

Consider a random variable,X,with Uniform density over the range 0 to 1.  Given that X is at least 0.4,find the probability that X is less than 0.6.
A)0.33
B)0.60
C)0.40
D)0.67
E)0.00

Accepted Answer

Given that $X$ is at least $0.4$, we know that the density of $X$ is 1 for $0.4\leq X\leq 1$. We need to find the probability that $X$ is less than $0.6$ given that $X\geq 0.4$.

This is equivalent to finding $P(0.4\leq X<0.6)$ divided by $P(X\geq 0.4)$.

$P(0.4\leq X<0.6)$ is the area of a rectangle with base $0.2$ and height $1$, which is $0.2$.

$P(X\geq 0.4)$ is the area of a rectangle with base $0.6$ and height $1$, which is $0.6$.

Therefore, the probability that $X$ is less than $0.6$ given that $X\geq 0.4$ is $\dfrac{0.2}{0.6} = 0.33$.

Hence, the answer is A) 0.33.

Question 3

The probability that a call received by a certain switchboard will be a wrong number is 0.02.Use the Poisson approximation to the binomial distribution to find the probability that among 140 calls received by the switchboard,there are at least two wrong numbers.
A)0.5305
B)0.7616
C)0.7689
D)0.2384
E)0.2311

Accepted Answer

Let X be the number of wrong numbers among 140 calls. We want to find P(X ≥ 2).

Since n = 140 is large and p = 0.02 is small, we can use the Poisson approximation to the binomial distribution:

X ~ Poisson(λ), where λ = np = 140 × 0.02 = 2.8.

Then,

P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
             = 1 - e^(-λ) - λe^(-λ)
             = 1 - e^(-2.8) - 2.8e^(-2.8)
             ≈ 0.7689

Therefore, the best choice is (C).

Question 4

An insurance company estimates that it should make an annual profit of $150 on each homeowner's policy written,with a standard deviation of $6700.If it writes 12,469 of these policies,what is the probability that the company will be profitable? Assume that policies are independent of each other and that the company's total profit follows a Normal model.
A)0.977
B)0.994
C)0.989
D)0.011
E)0.006

Accepted Answer

Given that the insurance company estimates an annual profit of $150 per homeowner's policy with a standard deviation of $6700, and it writes 12,469 policies, we can calculate the probability of profitability using the Central Limit Theorem. The mean total profit ($\mu_{total}$) is the number of policies (12,469) multiplied by the profit per policy ($150), and the standard deviation of the total profit ($\sigma_{total}$) is the standard deviation per policy ($6700) multiplied by the square root of the number of policies ($\sqrt{12469}$). The question asks for the probability that the company will be profitable, which means we are looking for P(X > 0).Using the Central Limit Theorem, we find that the distribution of the total profit is approximately normal. To find the probability of being profitable, we calculate the Z-score for a profit of $0 and then use the standard normal distribution to find the corresponding probability.$$Z = \frac{X - \mu_{total}}{\sigma_{total}}$$Given that the mean and standard deviation are based on the total number of policies, we can calculate the Z-score for $0 profit and then find the corresponding probability from the standard normal distribution. The calculation will show that the probability of the company being profitable (i.e., having a total profit greater than $0) is very high, which corresponds to choice B) 0.994. This high probability is due to the large number of policies written, which, according to the Central Limit Theorem, results in the total profit distribution being closely approximated by a normal distribution centered around a positive mean profit.

Question 5

Consider a random variable,X,with Uniform density over the range 0 to 1.  What is P(0.2 < X < 0.7)?
A)1.00
B)0.90
C)0.14
D)0.50
E)0.00

Accepted Answer

The answer of Consider a random variable,X,with Uniform density over...

Question 6

The probability that a car will have a flat tire while driving through a certain tunnel is 0.00006.Use the Poisson approximation to the binomial distribution to find the probability that among 13,000 cars passing through this tunnel,at most two will have a flat tire.
A)0.1394
B)0.1840
C)0.9554
D)0.8160
E)0.8606

Accepted Answer

The Poisson approximation to the binomial distribution can be used with λ = np, where n is the number of trials (13,000 cars) and p is the probability of success (0.00006). Thus, λ = 13,000 * 0.00006 = 0.78. The probability of at most 2 cars having a flat tire is given by P(X ≤ 2) = P(0) + P(1) + P(2), where P(k) = (e^(-λ) * λ^k) / k!. Substituting λ = 0.78, we calculate P(0), P(1), and P(2), and sum them up to find the probability, which is approximately 0.9554.

Question 7

In the 4 × 100 relay event,each of four runners runs 100 metres.A college team is preparing for a competition.The means and standard deviations of the times (in seconds)of their four runners are shown in the table: $$\begin{array} { c | c | c } 
\text { Runner } & \text { Mean } & \text { SD } \
\hline 1 & 12.01 & 0.10 \
2 & 12.26 & 0.15 \
3 & 11.93 & 0.10 \
4 & 11.81 & 0.08
\end{array}$$ Find the probability that in the competition they will beat their previous best time of 47.501 sec.Assume that the runners' performances are independent and that each runner's time follows a Normal model.
A)0.036
B)0.982
C)0.018
D)0.989
E)0.011

Accepted Answer

The answer of In the 4 × 100 relay event,each...

Question 8

The amount of money that Maria earns in a week is a random variable with a mean of $980 and a standard deviation of $25.The amount of money that Elena earns in a week is a random variable with a mean of $800 and a standard deviation of $10. You would like to use a Normal model to determine the probability that Maria's weekly income is at least $233.85 more than Elena's weekly income (the probability that the difference $M - E$ is at least $233.85).
Which of the following assumptions are needed?
A: Maria's weekly earnings are independent of Elena's weekly earnings.
B: Maria's weekly earnings and Elena's weekly earnings follow a Normal model.
C: Maria's weekly earnings are greater than Elena's weekly earnings
A)A and C
B)A only
C)B only
D)A,B,and C
E)A and B

Accepted Answer

We need assumption A to ensure that the two earnings are independent, and assumption B to use a Normal model for their distribution. Assumption C is not needed, as we are only interested in the difference between the earnings, not their individual values.

Question 9

The probability that a car will have a flat tire while driving through a certain tunnel is 0.00008.Use the Poisson approximation to the binomial distribution to find the probability that among 16,000 cars passing through this tunnel,at least one will have a flat tire.
A)0.7220
B)0.6441
C)0.2780
D)1.6462
E)0.3559

Accepted Answer

The Poisson approximation to the binomial distribution can be used with a mean (λ) equal to np, where n is the number of trials (16,000 cars) and p is the probability of success (0.00008). Thus, λ = 16,000 * 0.00008 = 1.28. The probability of at least one car having a flat tire is 1 minus the probability of no cars having a flat tire, which is calculated using the Poisson formula: P(X ≥ 1) = 1 - P(X = 0) = 1 - (e^(-λ) * λ^0 / 0!) = 1 - e^(-1.28) ≈ 0.7220.

Question 10

At a furniture factory,tables must be assembled,finished,and packaged before they can be shipped to stores.Based on past experience,the manager finds that the means and standard deviations (in minutes)of the times for each phase are as shown in the table: $$\begin{array} { l | l | l } 
\text { Phase } & \text { Mean } & \text { SD } \
\hline \text { Assembly } & 26.5 & 2.6 \
\text { Finishing } & 35.5 & 3.2 \
\text { Packaging } & 16.0 & 2.3
\end{array}$$ Find the probability that a table can be prepared for shipping in less than 67.61 minutes.Assume that the times for each phase are independent and that the times for each phase follow a Normal model.
A)0.014
B)0.023
C)0.986
D)0.045
E)0.955

Accepted Answer

The answer of At a furniture factory,tables must be assembled,finished,and...

Question 11

Consider a random variable,X,with Uniform density over the range 0 to 3.What is P(1.2 < X < 1.8)?
A)1.00
B)0.20
C)0.60
D)0.40
E)0.50

Accepted Answer

For a uniform distribution over the range 0 to 3, the total range is 3 - 0 = 3. The probability density function is constant over this range, so the probability of an event occurring within a subinterval of the range is simply the length of that subinterval divided by the total length of the range. Here, the subinterval is from 1.2 to 1.8, which has a length of 1.8 - 1.2 = 0.6. Therefore, the probability is 0.6 / 3 = 0.20.

Question 12

The rate of defects among CD players of a certain brand is 1.3%.Use the Poisson approximation to the binomial distribution to find the probability that among 100 such CD players received by a store,there are exactly three defectives.
A)0.1497
B)0.0748
C)0.0998
D)0.9002
E)0.1996

Accepted Answer

The answer of The rate of defects among CD players...

Question 13

Miguel buys a large bottle and a small bottle of juice.The amount of juice that the manufacturer puts in the large bottle is a random variable with a mean of 1018 ml and a standard deviation of 9 ml.The amount of juice that the manufacturer puts in the small bottle is a random variable with a mean of 506 ml and a standard deviation of 3 ml.Let the random variable T represent the total amount of juice in the two bottles.Suppose that you use a Normal model to determine the probability that the total amount of juice,T,in the two bottles is more than 1543.0 ml.Which of the following assumptions are needed to find the probability in this way? A: The amount of juice in the large bottle is independent of the amount of juice in the small bottle.
B: The amount of juice in each bottle follows a Normal model.
C: The amount of juice in the large bottle is greater than the amount of juice in the small bottle.
A)A and B
B)B only
C)A,B,and C
D)A only
E)No assumptions are needed.

Accepted Answer

To calculate the probability that the total amount of juice exceeds a certain value using a Normal model, it is necessary to assume that the amounts in the large and small bottles are independent (A) and that each follows a Normal distribution (B). Independence is required to correctly aggregate the means and variances, and the Normal distribution assumption allows for the use of Normal probability models. Assumption C is not required for the calculation and does not affect the statistical model.

Question 14

Sue buys a large packet of rice.The amount of rice that the manufacturer puts in the packet is a random variable with a mean of 1016 g and a standard deviation of 8 g.The amount of rice that Sue uses in a week has a mean of 200 g and a standard deviation of 29 g.If the weight of the rice remaining in the packet after a week can be described by a Normal model,what is the probability that the packet still contains more than 831.0 g after a week?
A)0.655
B)0.274
C)0.691
D)0.309
E)0.345

Accepted Answer

The answer of Sue buys a large packet of rice.The...

Question 15

The amount of money that Maria earns in a week is a random variable with a mean of $930 and a standard deviation of $30.The amount of money that Elena earns in a week is a random variable with a mean of $780 and a standard deviation of $15.If the difference between Maria's weekly income and Elena's weekly income can be described be a Normal model,what is the probability that Maria's weekly income is at least $223.79 more than Elena's weekly income? (In other words,what is the probability that the difference $\text { M-E }$ is at least $223.79?)Assume that Maria's earnings are independent of Elena's earnings.
A)0.018
B)0.986
C)0.011
D)0.982
E)0.014

Accepted Answer

The answer of The amount of money that Maria earns...

Question 16

Miguel buys a large bottle and a small bottle of juice.The amount of juice that the manufacturer puts in the large bottle is a random variable with a mean of 1024 ml and a standard deviation of 12 ml.The amount of juice that the manufacturer puts in the small bottle is a random variable with a mean of 504 ml and a standard deviation of 2 ml.If the total amount of juice in the two bottles can be described by a Normal model,what is the probability that the total amount of juice in the two bottles is more than 1554.8 ml?
A)0.014
B)0.011
C)0.018
D)0.986
E)0.982

Accepted Answer

The total amount of juice in the two bottles is a random variable with a mean of 1024 + 504 = 1528 ml and a standard deviation of sqrt(12^2 + 2^2) = 12.2 ml. The probability that the total amount of juice in the two bottles is more than 1554.8 ml is P(X > 1554.8) = P(Z > (1554.8 - 1528) / 12.2) = P(Z > 2.2) = 0.014.

Question 17

The rate of defects among CD players of a certain brand is 1.1%.Use the Poisson approximation to the binomial distribution to find the probability that among 270 such CD players received by a store,there is at most one defective.
A)0.1524
B)0.8476
C)0.2037
D)0.9487
E)0.7963

Accepted Answer

We are given that the rate of defects is 1.1%, which means the probability of a CD player being defective is 0.011. Let X be the number of defective CD players among 270. We want to find P(X ≤ 1).

Since the probability of a CD player being defective is small (0.011), we can approximate the binomial distribution with a Poisson distribution. The Poisson parameter λ is equal to the mean of the binomial distribution, which is n times the probability of success (defect) in one trial. Therefore, λ = 270 × 0.011 = 2.97.

Now we can use the Poisson distribution to find P(X ≤ 1):

P(X ≤ 1) = P(X = 0) + P(X = 1)
          = e^(-λ) (λ^0/0!) + e^(-λ) (λ^1/1!)
          = e^(-2.97) (1 + 2.97)
          = 0.2037

Therefore, the probability that among 270 CD players there is at most one defective is 0.2037, which is closest to choice C.

Question 18

Consider a random variable,X,with Uniform density over the range 0 to 1.  What is P(0.2 < X < 0.5 or 0.4 < X < 0.6)?
A)1.00
B)0.40
C)0.10
D)0.50
E)0.00

Accepted Answer

The answer of Consider a random variable,X,with Uniform density over...

Question 19

The probability that a call received by a certain switchboard will be a wrong number is 0.02.Use the Poisson approximation to the binomial distribution to find the probability that among 140 calls received by the switchboard,there are no wrong numbers.
A)0.9392
B)0.0608
C)0.1703
D)0.8297
E)0.0669

Accepted Answer

The answer of The probability that a call received by...

Question 20

Consider a random variable,X,with Uniform density over the range 0 to 1.  What is P(0.2 < X < 0.5 and 0.4 < X < 0.6)?
A)1.00
B)0.40
C)0.10
D)0.50
E)0.00

Accepted Answer

Since the density of X is uniform over the range 0 to 1, the probability of X being between 0.2 and 0.5 is (0.5-0.2)/1 = 0.3 and the probability of X being between 0.4 and 0.6 is (0.6-0.4)/1 = 0.2. The intersection of these two events is between 0.4 and 0.5, which has a probability of (0.5-0.4)/1 = 0.1. Therefore, P(0.2 < X < 0.5 and 0.4 < X < 0.6) = P(0.4 < X < 0.5) = 0.1. The answer is C.

Quiz 14: Random Variables

A slot machine at a casino pays out an average of $0.90,with a standard deviation of $120.It costs a dollar to play.If gamblers play this machine 6,969,600 times in a month,what is the probability that the casino will come out ahead? Assume that the casino's total profit follows a Normal model.

Consider a random variable,X,with Uniform density over the range 0 to 1. Given that X is at least 0.4,find the probability that X is less than 0.6.

The probability that a call received by a certain switchboard will be a wrong number is 0.02.Use the Poisson approximation to the binomial distribution to find the probability that among 140 calls received by the switchboard,there are at least two wrong numbers.

Consider a random variable,X,with Uniform density over the range 0 to 1. What is P(0.2 < X < 0.7) ?

The probability that a car will have a flat tire while driving through a certain tunnel is 0.00006.Use the Poisson approximation to the binomial distribution to find the probability that among 13,000 cars passing through this tunnel,at most two will have a flat tire.

The probability that a car will have a flat tire while driving through a certain tunnel is 0.00008.Use the Poisson approximation to the binomial distribution to find the probability that among 16,000 cars passing through this tunnel,at least one will have a flat tire.

Consider a random variable,X,with Uniform density over the range 0 to 3.What is P(1.2 < X < 1.8) ?

The rate of defects among CD players of a certain brand is 1.3%.Use the Poisson approximation to the binomial distribution to find the probability that among 100 such CD players received by a store,there are exactly three defectives.

The rate of defects among CD players of a certain brand is 1.1%.Use the Poisson approximation to the binomial distribution to find the probability that among 270 such CD players received by a store,there is at most one defective.

Consider a random variable,X,with Uniform density over the range 0 to 1. What is P(0.2 < X < 0.5 or 0.4 < X < 0.6) ?

The probability that a call received by a certain switchboard will be a wrong number is 0.02.Use the Poisson approximation to the binomial distribution to find the probability that among 140 calls received by the switchboard,there are no wrong numbers.

Consider a random variable,X,with Uniform density over the range 0 to 1. What is P(0.2 < X < 0.5 and 0.4 < X < 0.6) ?