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vova2212
10 days ago
7

Suppose the dealer incentive per vehicle for honda's acura brand in 2012 is thought to be bell-shaped and symmetrical with a mea

n of $2500 anda standard deviation of $300. Based in this information,
what interval of dealer incentives would we expect approximately 99.7% of vehicles to fall within?
Mathematics
1 answer:
lawyer [9.2K]10 days ago
8 0

Answer:

Between $1600 and $3400.

Step-by-step explanation:

According to the Empirical Rule, for a normally distributed random variable:

68% of values are found within 1 standard deviation of the mean.

95% of values reside within 2 standard deviations of the mean.

99.7% of values lie within 3 standard deviations of the mean.

Given data indicates:

Mean = 2500

Standard deviation = 300

What range of dealer incentives would accommodate approximately 99.7% of vehicles?

By applying the Empirical Rule, 99.7% are found within 3 standard deviations from the mean. Therefore,[[@TAG_34]]From 2500 - 3*300 = 1600 to 2500 + 3*300 = 3400.

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The probability that a randomly selected person has high blood pressure (the event H) is P(H) = 0.3 and the probability that a r
Svet_ta [9500]

Answer:

The likelihood of selecting an individual with high blood pressure who is also a runner is 2/3.

Step-by-step explanation:

Based on the data provided:

P( Choosing someone with high blood pressure )  =  0.3

or stated as P(H)  = 0.3

P( Choosing a runner )  =  0.4

or stated as P(R)  = 0.4

Now, P( Selecting a person with high blood pressure as well as being a runner)  = 0.2

⇒ P(H∩ R)  = 0.2

Next, we want to calculate the P(the selected individual is a runner, given they have high blood pressure)

which means we need to determine: P(R/H)

Using BAYES THEOREM:

P(R/H) = \frac{P(R\cap H)}{P(H)}

\implies P(R/H) = \frac{0.2}{0.3 } = \frac{2}{3}

So, the chance of choosing a person with high blood pressure who is a runner stands at 2/3.

4 0
26 days ago
Shalise competed in a jigsaw puzzle competition where participants are timed on how long they take to complete puzzles of variou
Svet_ta [9500]

Answer:

The small puzzle took 16% of the total time while the large puzzle took 84%.

Step-by-step explanation:

3 0
18 days ago
The Multiplication Rule says that P(A and B) = P(A) × P(B). What must be true about events A and B for this rule to apply?
Svet_ta [9500]

Response:

Events A and B exhibit independence.

Detailed explanation:

We are provided with

The multiplication principle applicable to events A and B

P(A\cap B)=P(A)\times P(B)

A correct assertion regarding events A and B is necessary for this principle to hold.

It is understood that for two events A and B to be independent

Consequently, the probability

P(A\cap B)=P(A)\cdot P(B)

Thus, the correct assertion concerning events A and B for the application of this principle is stated as

Events A and B are independent.

7 0
13 days ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
babunello [8412]

The 99% confidence interval for the actual mean difference between average mail-order and internet purchase amounts falls within [$(-31.82), $12.02].

Step-by-step clarification:

We know a random sample of 16 mail-order sales receipts shows a mean sale amount of $74.50 and a standard deviation of $17.25.

For internet sales, a random sample of 9 receipts gives a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal value utilized for constructing a 99% confidence interval for the true mean difference is given by;

                      P.Q.  =  

 ~

where,

= sample mean of mail-order sales = $74.50 \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

= sample mean of internet sales = $84.40

\bar X_1 = standard deviation for mail-order sales = $17.25

\bar X_2 = standard deviation for internet sales = $21.25

s_1 = number of mail-order sales receipts = 16

= number of internet sales receipts = 9s_2

Furthermore,  

 =  n_1 = 18.74

n_2

The actual mean difference between average mail-order and internet purchases is denoted by (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} }

).

Thus, the 99% confidence interval for (\mu_1-\mu_2) is expressed as;

      = \mu_1-\mu_2 Here, the t critical value at the 0.5% significance level with 23 degrees of freedom is 2.807.           =

          = [$-31.82, $12.02](\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

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

(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

5 0
14 days ago
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