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Morgarella [4.7K]

4 years ago

9

Assume that random guesses are made for seven multiple choice questions on an SAT test, so that there are n=7 trials, each wit

h probability of success (correct) given by p= 0.2. Find the indicated probability for the number of correct answers.
Find the probability that the number x of correct answers is fewer than 4.

1 answer:

Nezavi [6.7K]4 years ago

7 0

Answer:

Step-by-step explanation:

Let x be a random variable representing the number of guesses made for the sat questions.

Since the probability of getting the correct answer to a question is fixed for any number of trials and the outcome is either getting it correctly or not, then it is a binomial distribution. The probability of success, p = 0.2

Probability of failure, q = 1 - p = 1 - 0.2 = 0.8

the probability that the number x of correct answers is fewer than 4 is expressed as

P(x < 4)

From the binomial distribution calculator,

P(x < 4) = 0.97

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Write the ratio as a fraction in simplest form 25 to 45

brilliants [131]

Answer:

5/9

Step-by-step explanation:

3 0

3 years ago

There are 15 pieces of fruit in a bowl and 6 of them are apples? What percentage are apples

boyakko [2]

Answer:

40%

Step-by-step explanation:

You can first reduce this fraction by dividing both the numerator and denominator by the Greatest Common Factor of 6 and 15 using

GCF(6,15) = 3

6÷315÷3=25

We know that

25

is the same as

2÷5

Then using

Long Division for 2 divided by 5

gives us

0.4

Converting our number to a percentage:

0.4×100

=40%

5 0

3 years ago

HELP HELP HELP PLZZ!!!

kipiarov [429]

Answer:

Yes he is

Step-by-step explanation:

he is saying 1/4 and 2/8 of the coins are dimes 2/8 simplified is 1/4

let me know if this helps

3 0

4 years ago

Read 2 more answers

GIVING BRAINLIEST IF CORRECT PLEASE HELP ASA[P

Vitek1552 [10]

Answer:

A) b= 8 students

B) 1) 19/55 2) 28/55

Step-by-step explanation:

I'm sorry if I'm getting this whole concept wrong cause it felt way too easy to be right so I'm sorry if you get it wrong.

The reason the first question is 8 is because of the fact that you first have to add the students charted on the Venn-diagram and subtract that from 55, getting you to 8 students.

For the second one, the first part would be 19/55 becuase in the Venn-diagram it is shown that only 19 people passed both and so that be the numerator and the denominator would be the total of students, so it would be 55.

Same thing for part two except the fact that you have to add the students who only passed one subject, so combine 17 and 11, getting you to 28, thus getting the answer 28/55.

3 0

3 years ago