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inn [45]
3 years ago
8

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters tha

t are normally distributed with mean 3 cm and standard deviation 0.10 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.04 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm
What is the probability that the first machine produces an acceptable cork? (Round your answer to four decimal places.)
What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.)
Which machine is more likely to produce an acceptable cork? the first machine the second machine
Mathematics
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

0.6826 = 68.26% probability that the first machine produces an acceptable cork.

0.933 = 93.3% probability that the second machine produces an acceptable cork.

The second machine is more likely to produce an acceptable cork.

Step-by-step explanation:

When the distribution is normal, we use 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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

What is the probability that the first machine produces an acceptable cork?

The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.10 cm, which means that \mu = 3, \sigma = 0.1

Acceptable between 2.9 and 3.1, which means that this probability is the pvalue of Z when X = 3.1 subtracted by the pvalue of Z when X = 2.9.

X = 3.1

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

Z = \frac{3.1 - 3}{0.1}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 2.9

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

Z = \frac{2.9 - 3}{0.1}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

0.6826 = 68.26% probability that the first machine produces an acceptable cork.

What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.)

For the second machine, we have that \mu = 3.04, \sigma = 0.04. Same probability we have to find out. So

X = 3.1

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

Z = \frac{3.1 - 3.04}{0.04}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

X = 2.9

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

Z = \frac{2.9 - 3.04}{0.04}

Z = -3.5

Z = -3.5 has a pvalue of 0.0002

0.9332 - 0.0002 = 0.933

0.933 = 93.3% probability that the second machine produces an acceptable cork.

Which machine is more likely to produce an acceptable cork?

Second one has a higer probability, so it is more likely to produce an acceptable cork.

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Answer:

1. H0: P1 = P2

2. Ha: P1 ≠ P2

3. pooled proportion p = 0.542

4. P-value = 0.0171

5. The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

6. The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

Step-by-step explanation:

We should perform a hypothesis test on the difference of proportions.

As we want to test if there is significant difference, the hypothesis are:

Null hypothesis: there is no significant difference between the proportions (p1-p2 = 0).

Alternative hypothesis: there is significant difference between the proportions (p1-p2 ≠ 0).

The sample 1 (science), of size n1=135 has a proportion of p1=0.607.

p_1=X_1/n_1=82/135=0.607

The sample 2 (math), of size n2=92 has a proportion of p2=0.446.

p_2=X_2/n_2=41/92=0.446

The difference between proportions is (p1-p2)=0.162.

p_d=p_1-p_2=0.607-0.446=0.162

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{82+41}{135+92}=\dfrac{123}{227}=0.542

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.542*0.458}{135}+\dfrac{0.542*0.458}{92}}\\\\\\s_{p1-p2}=\sqrt{0.001839+0.002698}=\sqrt{0.004537}=0.067

Then, we can calculate the z-statistic as:

z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.162-0}{0.067}=\dfrac{0.162}{0.067}=2.4014

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

\text{P-value}=2\cdot P(z>2.4014)=0.0171

As the P-value (0.0171) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

We want to calculate the bounds of a 99% confidence interval of the difference between proportions.

For a 99% CI, the critical value for z is z=2.576.

The margin of error is:

MOE=z \cdot s_{p1-p2}=2.576\cdot 0.067=0.1735

Then, the lower and upper bounds of the confidence interval are:

LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.162-0.1735=-0.012\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.162+0.1735=0.335

The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

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