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ollegr [7]
1 year ago
10

Among 21- to 25-year-olds, 29% say they have driven while under the influence of alcohol. Suppose that three 21- to 25-year-olds

are selected at random.
a. What is the probability that at least 1 has driven while under the influence of alcohol?
b. Among 21- to 25-year-olds, 29% say they have driven while under the influence of alcohol.
Mathematics
1 answer:
Vilka [71]1 year ago
4 0

Answer:

0.6421

Step-by-step explanation:

In this case we have 3 trials and we have 2 options for each one. The driver has or hasn't been under alcohol influence. The probability that the driver has is 0.29 and the probabiility that the driver hasn't is 1 - 0.29 = 0.71

each trial is independent because we are assuming that the population of drivers in between 21 and 25 years old is very big.

The probability that one of them was under alcohol influence can be found by finding the probability that non of them was under alcohol influence because:

1 = p(x = 0) + p(x ≥ 1)

p(x ≥ 1) = 1 - p(0)

The probability that none of them was under alcohol influence is going to be:

0.71×0.71×0.71 = 0.3579

The probability of finding at least one driver that has been under alcohol influence is:

0.6421

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Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

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                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

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