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1 year ago

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Abdullah is a quality control expert at a factory that paints car parts. He knows that 20\ , percent of parts have an error in t

heir painting. He recommended a change in the painting process, and he wanted to see if this error rate had changed. He took a random sample of 400400400 parts painted in a month and found that 606060 had an error.

1 answer:

-Dominant- [34]1 year ago

3 0

The error rate has decreased after changing the painting process.

<u>Step-by-step explanation:</u>

Abdulla knows that 20 percent of the parts have an error in their painting. After suggesting changes in painting process, he wants to know whether the error rate has changed.

Number of parts in the random sample=400400400

Number of parts that had an error=606060

We have to determine what percentage of 400400400 is 606060

$606060=x/100 \times 400400400\\=0.15%$

After changing the painting process 0.15% of parts have error.

The previous percentage was 20.Hence the error rate has clearly changed.

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1. & 2. In the diagram below, points A, B, and C are collinear. Answer each of the following questions. The figure shown bel

KiRa [710]

Answer:

a). AB = 8 in

b). AB = 9.75 in

c). AC = 6.5 in

d). BC = 1.5 in

Step-by-step explanation:

a). Since, AB = AC + CB

Length of AC = 5 in. and CB = 3 in.

Therefore, AB = 5 + 3 = 8 in.

b). Given : AC = 6.25 in and CB = 3.5 in

Therefore, AB = AC + CB = 6.25 + 3.5

AB = 9.75 in.

c). Given: AB = 10.2 in. and BC = 3.7 in.

AB = AC + BC

AC = AB - BC

AC = 10.2 - 3.7

AC = 6.5 in

d). Given: AB = 4.75 in and AC = 3.25 in.

BC = AB - AC

BC = 4.75 - 3.25 = 1.5 in.

4 0

1 year ago

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so $n = 1210$ .

The mean of the sample is 1030.

The probability of a success is $\pi = \frac{1030}{1210} = 0.8512$ .

The standard deviation of the sample is $s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38$

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when $X = 1030$ .

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

$\mu = 0.82(1210) = 992.2$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1030 - 992.2}{12.38}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.This means that the county result is unusually high.

4 0

2 years ago

Monica has a bag of marbles. There are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick

natima [27]

Answer:

$\frac{7}{64}\approx 0.11$

Step-by-step explanation:

We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.

Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.

Since the probability of two independent compound events is always the product of probabilities of both events.

$P(\text{A and B})=P(A)*P(B)$

Now let us find probability of picking a red marble out of 16 (4+7+5) marbles.

$P(Red)=\frac{\text{Total red marbles}}{\text{Total marbles}}$

$P(Red)=\frac{4}{16}$

Probability of picking blue ball out of 16 (4+7+5) marbles:

$P(Blue)=\frac{\text{Total blue marbles}}{\text{Total marbles}}$

$P(Blue)=\frac{7}{16}$

Now let us find probability of Monica picking a red and then a blue marble.

$P(\text{Red and Blue})=\frac{4}{16}\times \frac{7}{16}$

$P(\text{Red and Blue})=\frac{1}{4}\times \frac{7}{16}$

$P(\text{Red and Blue})=\frac{7}{4*16}$

$P(\text{Red and Blue})=\frac{7}{64}$

$P(\text{Red and Blue})=0.109375\approx 0.11$

Therefore, the probability of picking a red and then blue marble is $\frac{7}{64}\approx 0.11$ .

7 0

2 years ago

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0

2 years ago

There are 527 pencils 646 erasers 748 sharpeners there are to be put in seperate packets containing the same number of items fin

blsea [12.9K]

Step 1

<u>Find the Greatest Common Factor (GCF)</u>

the prime factorization of each number is equal to

$527=17*31\\ 646=2*17*19\\ 748=2^{2} *11*17$

The GCF is 17

That means that the maximum number of packets is 17 and the maximum number of the items possible in each packet is

Number of pencils=527/17=31 pencils in each packet

Number of erasers=646/17=38 erasers in each packet

Number of sharpeners=748/17=44 sharpeners in each packet

7 0

2 years ago