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Butoxors [25]
4 years ago
11

A customer from Cavallaro's Fruit Stand picks a sample of 5 oranges at random from a crate containing 75 oranges, of which 6 are

rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)
Mathematics
1 answer:
boyakko [2]4 years ago
7 0

Answer:

0.341 is the probability that the sample contains 1 or more rotten oranges.

Step-by-step explanation:

We are given the following information:

We treat rotten as a success.

Number of oranges = 75

Number of rotten orange = 6

P(Rotten orange) =

=\dfrac{6}{75} = 0.08

Then the number of rotten oranges follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 5

We have to evaluate:

P(x \geq 1) =1 - P(x = 0)\\\\=1- \binom{5}{0}(0.08)^0(1-0.08)^5\\\\= 1 - 0.659\\= 0.341

0.341 is the probability that the sample contains 1 or more rotten oranges.

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Answer:

(a) Between 27 and 31 pounds per month = 0.62465

(b) More than 30.2 pounds per month = 0.1357

Step-by-step explanation:

We are given that each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds and the variable is approximately normally distributed.

<em>Let X = generation of newspaper for garbage or recycling</em>

The z-score probability distribution for normal distribution is given by;

             Z = \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population average = 28 pounds

            \sigma = population standard deviation = 2 pounds

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

(a) Probability of household generating between 27 and 31 pounds per month is given by = P(27 pounds < X < 31 pounds) = P(X < 31 pounds) - P(X \leq 27 pounds)

   P(X < 31 pounds) = P( \frac{X-\mu}{\sigma} < \frac{31-28}{2} ) = P(Z < 1.50) = 0.93319

  P(X \leq 27 pounds) = P( \frac{X-\mu}{\sigma} \leq \frac{27-28}{2} ) = P(Z \leq -0.50) = 1 - P(Z < 0.50)

                                                             = 1 - 0.69146 = 0.30854                       

<em>{Now, in the z table the P(Z  </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 and x = 0.50 in the z table which has an area of 0.93319 and 0.30854 respectively.}</em>

Therefore, P(27 pounds < X < 31 pounds) = 0.93319 - 0.30854 = 0.62465

(b) Probability of household generating more than 30.2 pounds per month is given by = P(X > 30.2 pounds)

   P(X > 30.2 pounds) = P( \frac{X-\mu}{\sigma} > \frac{30.2-28}{2} ) = P(Z > 1.10) = 1 - P(Z \leq<em> </em>1.10)

                                                                   = 1 - 0.8643 = 0.1357

<em>Now, in the z table the P(Z  </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.10 in the z table which has an area of 0.8643.</em>

3 0
4 years ago
Can someone help me with this pls, is super important
Alex17521 [72]
X = 50 degrees
y = 50 degrees

Explanation:

x + 130 = 180 (Angles between parallel lines)
x = 50

y = x = 50 (Corresponding angles)
7 0
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