Question

Underage drinking, Part II: We learned in Exercise 3.25 that about 69.7% of 18-20 year olds consumed alcoholic beverages in 2008. We now consider a random sample of fifty 18-20 year olds.(a) How many people would you expect to have consumed alcoholic beverages? (round to one decimal place) What is the standard deviation? (round to two decimal places)(b) Would you be surprised if there were 45 or more people who have consumed alcoholic beverages?Yes, 45 out of 50 is 90%No, it is just as likely as any other outcomeNo, 45 or more accounts for six different events -- this wouldn't be surprisingYes, 45 is more than two standard deviations above the expected value (mean)(c) What is the probability that 45 or more people in this sample have consumed alcoholic beverages? (round to four decimal places)

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: number of 18-20-year-olds that consume alcoholic beverages in a sample of 50.

The proportion of underage people that drinks are known to be p= 0.697

This variable is discrete. This experiment has two possible outcomes success or failure, we will call "success" each time we encounter an underage individual that consumes alcohol and "failure" will be counting an underage that does not consume alcohol. The number of repetitions of the trial is fixed n= 50. All randomly selected underage individuals are independent and the probability of success is constant trough the whole experiment p=0.697.

Then we can say that this variable has a binomial distribution and we will use that distribution to do the calculations.

a. Under a binomial distribution, the expected value is calculated as:

E(X)= n*p= 50*0.697= 34.85.

The variance of a binomial distribution is:

V(X)= n*p*(1-p)= 50*0.697*0.303= 10.55955

And the standard deviation is the square root of the variance:

√V(X)= 3.2495 ≅ 3.25

b. To know how rare the value 45, you have to see how distant it is concerning the expected value. For this you have to subtract the expected value and divide it by the standard deviation:

[X-E(X)]/√V(X)

(45-34.85)/3.25= 3.12

The value X=45 is 3.12 standard deviations above the mean, which means that it would be rare to find 45 people or more than consumed alcohol.

c. P(X≥45) = 1 - P(X<45)= 1 - P(X≤44)= 1 - 0.9994= 0.0006

I hope it helps!

Answer 2

The required values are:

a) Expect to have consumed alcoholic beverages$=np=34.85$

and, Standard Deviation $\sigma =3.25$

b) yes, 45 is more than two standard deviations above the expected value(mean)

C) Probability$=0.0015$

Standard Deviation and Probability:

• The standard deviation of a probability distribution is the degree of dispersion or the scatter of the probability distribution relative to its mean.
• It is the measure of the variation in the probability distribution from the mean.
• The standard deviation of a probability distribution is the square root of its variance.

a)

$n=50\\p=0.6970$

They expect to have consumed alcoholic beverages,

$=n \times p\\=50 \times 0.6970\\=34.85$

Standard Deviation,

$\sigma =\sqrt{\left ( np\left ( 1-p \right ) \right )} \\ =\sqrt{\left (34.85 \right )(1-0.6970)} \\ =3.25$

b) Yes, 45 is more than 2 standard deviations above the expected value (mean)

C) Probability,

$P=\left ( x > 44.5 \right ) \\ P=\frac{( Z > 44.5-34.85)}{3.25} \\ P=\left ( Z > 2.97 \right ) \\ P=1-P\left ( Z < 2.97 \right ) \\ P=1-0.9985 \\ P=0.0015$

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