##### General Statistics: Ch 5 Quiz

College: First year, College: Second year, College: Third year, College: Fourth year

The accompanying table describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods.

- Find the probability of getting exactly 7 peas with green pods.
- Find the probability of getting 7 or more peas with green pods.
- Which probability is relevant for determining whether 7 is an unusually high number of peas with green pods, the result from part (a) or part (b)?
- Is 7 an unusually high number of peas with green pods? Why or why not? Use 0.05 as the threshold for an unusual event.

**a.** 0.268

**b.** 0.364

*(0.268 + 0.096)*

**c.** The result from part (b)

**d.** No, since the appropriate probability is greater
than 0.05, it is not an unusually high number.

Based on data from a car bumper sticker study, when a car is randomly selected, the number of bumper stickers and the corresponding probabilities are as shown below.

- Does the given information describe a probability distribution?
- Assuming that a probability distribution is described, find its mean and standard deviation.
- Use the range rule of thumb to identify the range of values for usual numbers of bumper stickers.
- Is it unusual for a car to have more than one bumper sticker? Explain.

**a.** Yes

**b.** The mean is **0.6**.

The standard deviation is ** 1.5**.

**c.** The maximum usual value is **3.5**.

*max: μ + 2σ = 0.578 + 2(1.462161414) = 3.502322828*

The minimum usual value is ** 0**.

*min: μ – 2σ = 0.578 – 2(1.462161414) = -2.346322828*

**d.** No, because the probability of more than 1 bumper
sticker is 0.120, which is greater than 0.05.

*1 – (0.772 + 0.108) = 0.12*

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable.

- The number of free dash throw attempts before the first shot is made
- The number of light bulbs that burn out in the next week in a room with 18 bulbs
- The eye color of people on commercial aircraft flights
- The number of points scored during a basketball game
- The amount of snowfall in December in City A
- The number of people in a restaurant that has a capacity of 100

**a.** It is a discrete random variable.

**b.** It is a discrete random variable.

**c.** It is not a random variable.

**d.** It is a discrete random variable.

**e.** It is a continuous random variable.

**f.** It is a discrete random variable.

In the accompanying table, the random variable x represents the number of televisions in a household in a certain country.

Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation.

Its mean is **2.9**.

Its standard deviation is **1.3**.

Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.45. The table on the right describes the probability of having x number of girls.

Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation.

Is it unusual for a family of three children to consist of three girls?

* μ*

**= 1.35**

* σ*

**= 0.86**

No, because the probability of having 3 girls is greater than 0.05.

Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.31. The table on the right describes the probability of having x number of girls.

Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation.

Is it unusual for a family of three children to consist of three girls?

* μ* = 0.93

* σ* = 0.80

Yes, because the probability of having 3 girls is less than or equal to 0.05.

In a state's Pick 3 lottery game, you pay $1.45 to select a sequence of three digits (from 0 to 9), such as 422. If you select the same sequence of three digits that are drawn, you win and collect $439.13.

- How many different selections are possible?
- What is the probability of winning?
- If you win, what is your net profit?
- Find the expected value.
- If you bet $ 1.45 in a certain state's Pick 4 game, the expected value is negative $ 1.01. Which bet is better, a $1.45 bet in the Pick 3 game or a $ 1.45 bet in the Pick 4 game? Explain.

**a.** 1,000

**b.** 0.001

**c.** $437.68

*($439.13 – $1.45)*

**d.** $(1.01)

*(–$1.45 + $439.13/1000)*

*= -1.45 + 0.44*

*= -1.01*

**e.** Neither bet is better because both games have the
same expected value.

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder.

Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation.

* μ*

**= 2.9**

* σ*

**= 1.1**

Determine whether or not the procedure described below results in a binomial distribution. If it is not binomial, identify at least one requirement that is not satisfied.

Six hundred different voters in a region with two major political parties, A and B, are randomly selected from the population of 3.7 million registered voters. Each is asked if he or she is a member of political party A.

Yes, the result is a binomial probability distribution.

A certain TV show recently had a share of 85, meaning that among the TV sets in use, 85% were tuned to that show. Assume that an advertiser wants to verify that 85% share value by conducting its own survey, and a pilot survey begins with 8 households having TV sets in use at the time of the TV show broadcast.

- Find the probability that all of the households are tuned to the TV show.
- Find the probability that exactly 7 households are tuned to the TV show.
- Find the probability that at least 7 households are tuned to the TV show.
- If at least 7 households are tuned to the TV show, does it appear that the 85% share value is wrong? Why or why not?

**a.** The probability that all of the households are
tuned to the TV show is **0.272**.

*P(x=8) = (0.85) ^{8} = 0.272490525*

**b.** The probability that exactly 7 households are
tuned to the TV show is **0.385**.

*P(x=7) = ( ^{8} _{7} ) x (0.85)^{7} x (0.15)^{1}*

*= 0.3846925059*

**c.** The probability that at least 7 households are
tuned to the TV show is ** 0.657**.

*P(x ≥ 7) = P(x=7 or x=8)*

*= 0.27249 + 0.38469*

*= 0.65718*

**d.** No, because 7 households tuned to the TV show is
not unusually high if the share is 85%.

Refer to the accompanying technology display. The probabilities in the display were obtained using the values of n = 5 and p = 0.738. In a clinical test of a drug, 73.8% of the subjects treated with 10 mg of the drug experienced headaches. In each case, assume that 5 subjects are randomly selected and treated with 10 mg of the drug.

Find the probability that more than one subject experiences headaches.

Is it reasonable to expect that more than one subject will experience headaches?

The probability that more than one subject experiences headaches is
**0.9814**.

*P(x > 1) = P(x=2 or x=3 or ... or x=5)*

*= 1 – P(x=0 or x=1)*

*= 1 – (0.0012 + 0.0174)*

*= 0.9814*

**Yes, because the event that the number of subjects that
experience headaches is less than or equal to one is unlikely.**

Nine peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod.

Find the probability that among the 9 offspring peas, at least 8 have green pods.

Is it unusual to get at least 8 peas with green pods when 9 offspring peas are generated? Why or why not?

The probability that at least 8 of the 9 offspring peas have green
pods is **0.300**.

*P(x ≥ 8) = P(x=8 or x=9)*

*= [ ( ^{9} _{8} ) x
(0.75)^{8} x (0.25)^{1} ] +
[ (0.75)^{9} ]*

*= 0.225 + 0.075*

*= 0.3003387451*

**No, because the probability of this occurring is not small.**

A brand name has a 60% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 5 randomly selected consumers.

- What is the probability that exactly 4 of the selected consumers recognize the brand name?
- What is the probability that all of the selected consumers recognize the brand name?
- What is the probability that at least 4 of the selected consumers recognize the brand name?
- If 5 consumers are randomly selected, is 4 an unusually high number of consumers that recognize the brand name?

**a.** The probability that exactly 4 of the 5 consumers
recognize the brand name is **0.259**.

*P(x=4) = ( ^{5} _{4} ) x (0.60)^{4} x (0.40)^{1}*

*= 0.2592*

**b.** The probability that all of the selected consumers
recognize the brand name is **0.078**.

*P(x=5) = (0.60) ^{5}*

*= 0.07776*

**c.** The probability that at least 4 of the selected
consumers recognize the brand name is **0.337**.

*P(x ≥ 4) = P(x=4 or x=5)*

*= 0.259 + 0.078*

*= 0.33696*

**d.** No, because the probability that 4 or more of the
selected consumers recognize the brand name is greater than 0.05.

Assume that a procedure yields a binomial distribution with n = 2 trials and a probability of success of p = 0.50. Use a binomial probability table to find the probability that the number of successes x is exactly 1.

P(1) = **0.500**

Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial.

n=6, x=4, p=0.25

P(4) = **0.033**

*P(x=4) = ( ^{6} _{4} ) x (0.25)^{4} x (0.75)^{2}*

*= 0.0329589844*

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied.

Determining whether each of 50 mp3 players is acceptable or defective

Yes, because all 4 requirements are satisfied.

A TV show, *Lindsay and Tobias*, recently had a share of
15, meaning that among the TV sets in use, 15% were tuned to that
show. Assume that an advertiser wants to verify that 15% share value
by conducting its own survey, and a pilot survey begins with 18
households having TV sets in use at the time of a *Lindsay and
Tobias* broadcast.

- Find the probability
that none of the households are tuned to
*Lindsay and Tobias*. - Find the probability that at least one household
is tuned to
*Lindsay and Tobias*. - Find the
probability that at most one household is tuned to
*Lindsay and Tobias*. - If at most one household is tuned to Lindsay and Tobias, does it appear that the 15% share value is wrong? Why or why not?

**a.** **0.054**

*P(x=0) =
(0.85) ^{18} = 0.0536464098*

**b. 0.946**

*P(x ≥ 1) = P(x=1 or x=2 or ... or x=18)*

*= 1 – P(x=0)*

*= 1 – 0.054*

*= 0.9463535902*

**c. 0.224**

*P(x ≤ 1) = P(x=0 or x=1)*

*= 0.054 + [ (^{} ^{18}
_{1} ) x (0.15)^{1} x (0.75)^{17} ]*

*= 0.054 + 0.170*

*= 0.2240526527*

**d.** No, because with a 15% rate, the probability of
at most one household is greater than 0.05.

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied.

Treating 150 bald men with a special shampoo and asking them how their scalp feels

No, because there are more than two possible outcomes.

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied.

Recording the genders of 150 people in a statistics class

Yes, because all 4 requirements are satisfied.

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.

Use the given values of n and p to find the mean **μ**
and standard deviation **σ**.

Also, use the range rule of thumb to find the minimum usual value
**μ** – 2**σ** and the maximum usual value
**μ** + 2**σ**.

n = 1550, p = 3 / 5

**μ **= **930**

*
μ = np*

*= (1550)(0.60)*

*= 930*

**σ **= **19.3**

**σ** = √npq

= √(1550 x 0.60 x 0.40)

= 19.28730152

**μ** – 2**σ** = **891.4**

**μ** + 2** σ **=

**968.6**