Question

1. the average thickness of books on a library shelf is 8.3 centimeters. the standard deviation is 0.6 centimeter. if 20% of the books are oversized, find the minimum thickness of the oversized books on the library shelf. assume that the variable is normally distributed. 8.804 cm
2. the average repair cost of a microwave oven is $55, with a standard deviation of $8. the costs are normally distributed. if 12 ovens are repaired, find the probability that the mean of the repair bills will be greater than $60. 0.015
3. the average electric bill in a residential area is $72 for the month of april. the standard deviation is $6. if the amounts of the electric bill are normally distributed, find the probability that the mean of the bill for 15 residents will be less than $75. 0.9738

Explanation:

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Step1: Find the z - score for the 80th percentile

We know that if 20% of the books are oversized, then we want to find the value corresponding to the 80th percentile. Looking up in the standard - normal distribution table (z - table), the z - score \(z\) corresponding to a cumulative probability of 0.8 is approximately \(z = 0.84\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value we want to find, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. We know that \(\mu = 8.3\) cm, \(\sigma=0.6\) cm, and \(z = 0.84\). Rearranging the formula for \(x\) gives \(x=\mu + z\sigma\).
Substitute the values: \(x = 8.3+0.84\times0.6\)
\[

$$\begin{align*} x&=8.3 + 0.504\\ x&=8.804\text{ cm} \end{align*}$$

\]

Step1: Calculate the standard error of the mean

The standard error of the mean \(\sigma_{\bar{x}}\) for a sample of size \(n\) from a population with standard deviation \(\sigma\) is given by \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 8\), \(n = 12\), so \(\sigma_{\bar{x}}=\frac{8}{\sqrt{12}}\approx2.3094\).

Step2: Calculate the z - score

The z - score is calculated using the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), where \(\bar{x}=60\), \(\mu = 55\), and \(\sigma_{\bar{x}}\approx2.3094\). So \(z=\frac{60 - 55}{2.3094}\approx2.16\).

Step3: Find the probability

We want to find \(P(\bar{X}>60)\), which is equivalent to \(1 - P(\bar{X}\leq60)\). Looking up the z - score of \(z = 2.16\) in the standard - normal distribution table, \(P(Z\leq2.16)=0.9846\). So \(P(\bar{X}>60)=1 - 0.9846 = 0.0154\approx0.015\)

Step1: Calculate the standard error of the mean

The standard error of the mean \(\sigma_{\bar{x}}\) for a sample of size \(n\) from a population with standard deviation \(\sigma\) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 6\), \(n = 15\), so \(\sigma_{\bar{x}}=\frac{6}{\sqrt{15}}\approx1.5492\).

Step2: Calculate the z - score

The z - score is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), where \(\bar{x}=75\), \(\mu = 72\), and \(\sigma_{\bar{x}}\approx1.5492\). So \(z=\frac{75 - 72}{1.5492}\approx1.94\).

Step3: Find the probability

Looking up the z - score of \(z = 1.94\) in the standard - normal distribution table, \(P(Z\leq1.94)=0.9738\)

Answer:

\(8.804\) cm

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