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3 years ago

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A random sample of 12 college graduates revealed that they worked an average of 5.5 years on the job before being promoted. The

sample standard deviation was 1.1 years. Using the 95% level of confidence, what is the confidence interval for the population mean?

1 answer:

Hoochie [10]3 years ago

6 0

Answer:

90%

Step-by-step explanation:

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5x + 4 = -6

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5x=-10

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The value of the first expression is x=-2.

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The value of the second expression is x=-2.

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Question 4 of 10, Step 1 of 1 1/out of 10 Correct Certify Completion Icon Tries remaining:0 The Magazine Mass Marketing Company

erastovalidia [21]

Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

This means that $n = 16$

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that $p = 0.5$

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0$

$P(X = 1) = C_{16,1}.(0.5)^{1}.(0.5)^{15} = 0.0002$

$P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018$

$P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0002 + 0.0018 + 0.0085 = 0.0105$

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

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2 years ago