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UNO [17]
3 years ago
15

The Perfect Pumpkin Patch claims that their princess pumpkins have the mean weight of at least 12 lbs. You suspect they are embe

llishing the truth somewhat. Your plan is to bring a small child to find 40 pumpkins (at random, because small children are perfect random number generators) and weigh the pumpkins. How are you going to formulate your null and alternative hypotheses for this test of significance
Mathematics
1 answer:
ser-zykov [4K]3 years ago
5 0

Answer: Null hypothesis : \mu\geq12

Alternative hypothesis : \mu  

Step-by-step explanation:

  • Hypothesis is formed as per the researcher's objective about the population parameter.
  • Null hypothesis contains '=', '≤', '≥' , where as Alternative hypothesis  [opposite of null hypothesis] contains '≠', '<','>'.

Let \mu = mean weight princess pumpkins (in lbs).

Perfect Pumpkin Patch claims \mu\geq12.

i.e. Null hypothesis : \mu\geq12

Alternative hypothesis : \mu  

So the required null and alternative hypotheses :

Null hypothesis : \mu\geq12

Alternative hypothesis : \mu  

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A quiz consists of six multiple choice questions. Each question has four choices. A student who forgot to study guesses randomly
jeka94

Answer:

0.9945 = 99.45% probability that the student answers at most four questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student answers correctly, or he/she does not. The probability of the student answering a question correctly is independent of any other question. This 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.

A quiz consists of six multiple choice questions.

This means that n = 6

Each question has four choices. Student guesses randomly.

This means that p = \frac{1}{4}

What is the probability that the student answers at most four questions correctly?

This is:

P(X \leq 4) = 1 - P(X > 4)

In which

P(X > 4) = P(X = 5) + P(X = 6). So

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

P(X = 5) = C_{6,5}.(0.25)^{5}.(0.75)^{1} = 0.0053

P(X = 6) = C_{6,6}.(0.25)^{6}.(0.75)^{0} = 0.0002

P(X > 4) = P(X = 5) + P(X = 6) = 0.0053 + 0.0002 = 0.0055

P(X \leq 4) = 1 - P(X > 4) = 1 - 0.0055 = 0.9945

0.9945 = 99.45% probability that the student answers at most four questions correctly

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