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2 years ago

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A consumer advocacy group suspects that a local supermarket's 10 oz. packages of cheddar cheese actually weigh less than 10 oz.

The standard deviation is 0.15 oz. The group took a random sample of 35 such packages and found that the mean weight for the sample was 9.95 oz. Calculate the confidence interval. Don't forget to label. Please go 4 decimal spaces on the error, but 2 on the boundaries.

1 answer:

Maksim231197 [3]2 years ago

3 0

Answer: $(9.90,\ 10.00)$

Step-by-step explanation:

Confidence interval for population mean:

$\overline{x}\pm z^c\dfrac{\sigma}{\sqrt{n}}$

where , $\overline{x}$ = sample mean , $\sigma$ = population standard deviation, n= sample size, $z^c$ = Critical z-value.

As per given, $n= 35,\ \ \ \overline{x} =9.95,\ \ \ \ \sigma=0.15$

We will find 95% confidence interval for which $z^c=1.96$

Required confidence interval:

$9.95\pm(1.96)\dfrac{0.15}{\sqrt{35}} \\\\= 9.95\pm (0.049695)\\\\=( 9.95-0.049695,\ 9.95+0.049695)\\\\=(9.900305,\ 9.999695)\approx(9.90,\ 10.00)$

Required confidence interval : $(9.90,\ 10.00)$

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(a) $P(x\le 3) = 0.75$

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$\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

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$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

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Solving (b): Probability of at most 3 CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x \ge 5) = P(5) + P(6\ or\ more)$

This gives:

$P(x\ge 5) = 0.05 + 0.05$

$P(x \ge 5) = 0.10$

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$\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}$

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$P(x = 1\ or\ 2) = P(1) + P(2)$

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$\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

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$P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)$

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$P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05$

$P(x > 2) = 0.45$

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