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Tamiku [17]
3 years ago
12

Answer this question to get marked as barinliest!!!!!!

Mathematics
1 answer:
Shalnov [3]3 years ago
3 0

Answer:

2x to the 5th power

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Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on
FinnZ [79.3K]

The question is incomplete! Complete question along with answers and step by step explanation is provided below.

Question:

(a) Binomial probability distributions depend on the number of trials n of a binomial experiment and the probability of success p on each trial. Under what conditions is it appropriate to use a normal approximation to the binomial? (Select all that apply.)

nq > 10

np > 5

p > 0.5

np > 10

p < 0.5

nq > 5

(b) What is the probability of "12" or fewer successes for a binomial experiment with 20 trials. The probability of success on a single trial is 0.50. Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

Answer:

(a) The correct options are np > 5 and nq > 5

(b) P(x ≤ 12) = 0.8133

Step-by-step explanation:

Please refer to the attached images for explanation, I am unable to type in text editor due to some technical error!

8 0
3 years ago
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

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

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

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
3 years ago
Before Nina bought groceries on April 22, she had a balance of $487.25 in her checking account. Nina wrote her transactions in h
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Step-by-step explanation: i did it before

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3 years ago
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In an isosceles triangle, one of the equal interior angles is equal to the adjacent exterior
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Answer:

Step-by-step explanation:

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2 years ago
The length of a rectangle is 8 inches longer than half of its width.The area of the rectangle is 18 square inches.How many inche
SOVA2 [1]

Answer:

The correct answer is 2 inches.

Step-by-step explanation:

Let l inches and w inches be the length and width of a rectangle respectively.

According to the given problem, l - 8 = \frac{w}{2}.

Area of the rectangle is given to be, according to the question, 18 square inches.

Thus l × w = 18

⇒ l × (2l - 16) = 18

⇒ 2 l^{2} - 16l - 18 = 0

⇒  l^{2} - 8l - 9 = 0

⇒ ( l - 9) ( l + 1) = 0

The possible values of l are 9 and -1. As the length cannot be equal to -1, thus the value of the length is 9 inches.

Width of the rectangle is 2 inches.

5 0
3 years ago
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