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lawyer [7]
3 years ago
7

The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 oun

ces and a standard deviation of 0.20 ounce. The company receives complaints from consumers who actually measure the amount of soda in the cans and claim that the volume is less than the advertised 12 ounces. What percentage of the soda cans contain more than the advertised 12 ounces of soda
Mathematics
1 answer:
LiRa [457]3 years ago
7 0

Answer:

The correct answer is "0.9332".

Step-by-step explanation:

Given that:

\mu=12.30

\sigma=0.20

So,

= P(X>12)

= 1-P(X

= 1-P(\frac{X-\mu}{\sigma} < \frac{12-12.30}{0.20} )

= 1-P(Z

= 1-P(Z

= 1-0.0668

= 0.9332

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Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

(a)

The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

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For 5% level of significance the two tailed critical value of <em>z</em> is:

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As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

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P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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