A random sample of 100 units is drawn a production process every

half hour. The fraction of nonconforming product manufactured is

0.03. What is the probability that ˆp ≤ 0.04 if the fraction

nonconforming really is 0.03?

Answers

Answer 1

The probability that ˆp ≤ 0.04 if the fraction nonconforming really is 0.03 is written below.

What is a fraction?

A number that represents a rational number is called a common fraction.

X=0,1,2,3,…..; q=1-p

The probability that (p∧ ≤ 0.06) , substitute the values of sample units (n), and the probability of nonconformities (p) in the probability mass function of the binomial distribution.

Consider x   to be the number of non-conformities. It follows a binomial distribution, with n   being 50 and p  being 0.03. That is,

binomial (50,0.02)

Also, the estimate of the true probability is,

p∧  = x/50

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

The calculation is obtained as

P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)

        = 50cx ₓ (0.03)x ₓ (1-0.03)50-x  

=    (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ      (0.03)3 ₓ (1- 0.03)50-3 )

=(  ₓ (0.03)0 ₓ (1-0.03)50-0 +  ₓ (0.03)1 ₓ (1-0.03)50-1 +   ₓ (0.03)2 ₓ (1-0.03)50-2   ₓ (0.03)3 ₓ (1-0.03)50-3 )

Therefore, the probability is written above.

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Related Questions

Can someone help me with this statistics with explanation please

Answers

Answer:

b

Explanation:

The outcome 0, 1, 1 has a sample mean of x-bar = (0+1+1)/3 = 2/3 = 0.667

For independent outcomes, the probability of an outcome is the product of the probabilities of each individual event, which in this case would be the probability of getting a 0, the probability of getting a 1, and the probability of getting another 1.

If we assume that the probability of getting a 0, a 1, or a 1 is the same (let's say p) the probability of getting the outcome 0, 1, 1 would be ppp = p^3

So the answer is A. x-bar = 0.667, p = p^3

Note that the question doesn't provide the probability of getting a 0, 1, or 1, so I assumed the probability is the same.

Can you guys help me with 4 b)?

Answers

It’s b lol reeeeeeeeeeeeee

The probability that at least 20 of the novels have fewer than 400 pages is given as follows:

0.0823 = 8.23%.

How to obtain probabilities using the normal distribution?

The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].

The parameters for the binomial distribution are given as follows:

p = 0.3, n = 50.

Hence the mean and the standard deviation for the approximation are given as follows:

[tex]\mu = 0.3 \times 50 = 15[/tex][tex]\sigma = \sqrt{0.3 \times 0.7 \times 50} = 3.24[/tex]

Using continuity correction, the probability that at least 20 of the novels have fewer than 400 pages is P(x > 19.5), which is one subtracted by the p-value of Z when X = 19.5, hence:

Z = (19.5 - 15)/3.24

Z = 1.39

Z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823 = 8.23%.

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In a city library, the mean number of pages in a novel is 525 with a standard deviation of 200.
Furthermore, 30% of the novels have fewer than 400 pages. Suppose that you randomly select 50 nov from the library.

What is the probability that at least 20 of the novels have fewer than 400 pages?

Answers

The probability that at least 20 of the novels have fewer than 400 pages is 0.086.

To find this probability, we can use the cumulative distribution function (CDF) of a binomial distribution. The binomial distribution models the number of successful outcomes in a fixed number of independent trials, where each trial has a probability of success.

In this case, the number of trials is 50, the number of successful outcomes is the number of novels with fewer than 400 pages, and the probability of success is 0.3 (30% of the novels have fewer than 400 pages).

The CDF of the binomial distribution gives the probability of having at most a certain number of successful outcomes. To find the probability of having at least 20 successful outcomes, we can subtract the probability of having less than 20 successful outcomes from 1:

[tex]P(X > = 20) = 1 - P(X < 20)[/tex]

Where X is the number of successful outcomes.

Using a binomial calculator, we can find that  [tex]P(X > = 20) = 0.086[/tex]

So the probability that at least 20 of the novels have fewer than 400 pages is 0.086.

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