Mathematics, 04.07.2020 14:01 moneyshi68

What is the probability that between 970 and 990 of these intervals contain the corresponding value of μ? [Hint: Let Y = the number among the 1000 intervals that contain μ. What kind of random variable is Y?] (Round your answer to four decimal places.)

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Answer from: parapraxis

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

Question:

Consider the next 1000 98% CIs for μ that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of μ?

What is the probability that between 970 and 990 of these intervals contain the corresponding value of μ? [Hint: Let Y = the number among the 1000 intervals that contain μ. What kind of random variable is Y?]

(Round your answer to four decimal places.)

The expected value is

$E(Y) = n \times p \\\\E(Y) = 1000 \times 0.98 \\\\E(Y) = 980$

$P(970 < Y < 990) = 0.9753 \\\\P(970 < Y < 990) = 97.53 \%$

There is a 97.53% probability that between 970 and 990 of these intervals contain the corresponding value of μ.

Step-by-step explanation:

Consider the next 1000 98% CIs for μ that a statistical consultant will obtain for various clients.

From the above information, we know that,

n = 1000

p = 0.98

How many of these 1000 intervals do you expect to capture the corresponding value of μ?

The expected value is given by

$E(Y) = n \times p \\\\E(Y) = 1000 \times 0.98 \\\\E(Y) = 980$

What is the probability that between 970 and 990 of these intervals contain the corresponding value of μ?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and p is also greater than 0.50.

n×p ≥ 10

1000×0.98 ≥ 10

980 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

1000×(1 - 0.98) ≥ 10

20 ≥ 10 (satisfied)

Let Y be the random number among the 1000 intervals that contain μ

$P(970 < Y < 990) = P(Y < 990) - P(Y < 970) \\\\P(970 < Y < 990) = P( Z < \frac{Y - \mu}{\sigma}) - P( Z < \frac{Y - \mu}{\sigma} )\\\\$

Where the mean is

$\mu = 980$

and the standard deviation is

$\sigma = \sqrt{n \times p(1-p)} \\\\\sigma = \sqrt{1000 \times 0.98(1-0.98)} \\\\\sigma = 4.4272$

Finally, the required probability is

$P(970 < Y < 990) = P( Z < \frac{990 - 980}{4.4272}) - P( Z < \frac{970 - 980}{4.4272} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(970 < Y < 990) = P( Z < \frac{989.5 - 980}{4.4272}) - P( Z < \frac{969.5 - 980}{4.4272} )\\\\P(970 < Y < 990) = P( Z < \frac{9.5}{4.4272}) - P( Z < \frac{-10.5}{4.4272} )\\\\P(970 < Y < 990) = P( Z < 2.15) - P( Z < -2.37 )\\\\$

From the z-table, the z-score corresponding to 2.15 is 0.9842

From the z-table, the z-score corresponding to -2.37 is 0.0089

$P(970 < Y < 990) = 0.9842 - 0.0089\\\\P(970 < Y < 990) = 0.9753 \\\\P(970 < Y < 990) = 97.53 \%$

Therefore, there is a 97.53% probability that between 970 and 990 of these intervals contain the corresponding value of μ.

Answer from: Quest

i need more details

step-by-step explanation:

Answer from: Quest

The answer would be (8+7z)/5

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