Part 1:
The z-score of a value, a, given the mean, μ, and the standard deviation, σ, is given by
![z= \frac{a-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7Ba-%5Cmu%7D%7B%5Csigma%7D%20)
Thus, given that t<span>he
average cost per ounce for glass cleaner is 7.7 cents with a standard
deviation of 2.5 cents.
The z-score of glass cleaner with a cost
of 10.1 cents per ounce is given by
![z= \frac{10.1-7.7}{2.5} = \frac{2.4}{2.5} =\bold{0.96}](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B10.1-7.7%7D%7B2.5%7D%20%3D%20%5Cfrac%7B2.4%7D%7B2.5%7D%20%3D%5Cbold%7B0.96%7D)
Part 2:
Assuming the data is normally distribution, the probability that a normally distributed data is between two values a and b is given by
![P(a\ \textless \ X\ \textless \ b)=P(X\ \textless \ b)-P(X\ \textless \ a) \\ \\ =P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)](https://tex.z-dn.net/?f=P%28a%5C%20%5Ctextless%20%5C%20X%5C%20%5Ctextless%20%5C%20b%29%3DP%28X%5C%20%5Ctextless%20%5C%20b%29-P%28X%5C%20%5Ctextless%20%5C%20a%29%20%5C%5C%20%20%5C%5C%20%3DP%5Cleft%28z%5C%20%5Ctextless%20%5C%20%20%5Cfrac%7Bb-%5Cmu%7D%7B%5Csigma%7D%20%5Cright%29-P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%20%5Cfrac%7Ba-%5Cmu%7D%7B%5Csigma%7D%20%5Cright%29)
Thus, given that a</span><span> candy bar takes an average of 3.4 hours to move through an assembly line and that the standard deviation is 0.5 hours</span><span>.
The probability </span>that the candy bar will take between 3 and 4 hours to move through the line is given by
![P(3\ \textless \ X\ \textless \ 4)=P(X\ \textless \ 4)-P(X\ \textless \ 3) \\ \\ =P\left(z\ \textless \ \frac{4-3.4}{0.5} \right)-P\left(z\ \textless \ \frac{3-3.4}{0.5} \right)=P(z\ \textless \ 1.2)-P(z\ \textless \ -0.8) \\ \\ =0.88493-0.21186=\bold{0.6731}](https://tex.z-dn.net/?f=P%283%5C%20%5Ctextless%20%5C%20X%5C%20%5Ctextless%20%5C%204%29%3DP%28X%5C%20%5Ctextless%20%5C%204%29-P%28X%5C%20%5Ctextless%20%5C%203%29%20%5C%5C%20%5C%5C%20%3DP%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7B4-3.4%7D%7B0.5%7D%20%5Cright%29-P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7B3-3.4%7D%7B0.5%7D%20%5Cright%29%3DP%28z%5C%20%5Ctextless%20%5C%201.2%29-P%28z%5C%20%5Ctextless%20%5C%20-0.8%29%20%5C%5C%20%20%5C%5C%20%3D0.88493-0.21186%3D%5Cbold%7B0.6731%7D)
Part 3
Given that t<span>he
average noise level in a diner is 30 decibels with a standard deviation
of 4 decibels.
The value for which the noice level is below 99% of the time</span> is the value for which the probability that the noice level is below that value is 0.99.
<span>Assuming the data is normally distribution, the probability that a
normally distributed data is less than a value, a, is given by
![P(X\ \textless \ a)=P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)](https://tex.z-dn.net/?f=P%28X%5C%20%5Ctextless%20%5C%20a%29%3DP%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7Ba-%5Cmu%7D%7B%5Csigma%7D%20%5Cright%29)
Thus, given that </span>t<span>he
average noise level in a diner is 30 decibels with a standard deviation
of 4 decibels.
The value for which 99% of the time, the noise level is below is obtained as follows:
![P(X\ \textless \ a)=0.99 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{a-30}{4} \right)=P(z\ \textless \ 2.33) \\ \\ \Rightarrow\frac{a-30}{4}=2.33 \\ \\ \Rightarrow a=4(2.33)+30=\bold{39.32}](https://tex.z-dn.net/?f=P%28X%5C%20%5Ctextless%20%5C%20a%29%3D0.99%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20P%5Cleft%28z%5C%20%5Ctextless%20%5C%20%5Cfrac%7Ba-30%7D%7B4%7D%20%5Cright%29%3DP%28z%5C%20%5Ctextless%20%5C%202.33%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Cfrac%7Ba-30%7D%7B4%7D%3D2.33%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20a%3D4%282.33%29%2B30%3D%5Cbold%7B39.32%7D)
Note: there is standard deviation value was omitted in the question post but I used 4 as my standard deviation. So you can substitute the appropriate standard deviation for you question.
Part 4:
From the empirical rule, it is known that for a normally distributed data that 95% of the data are within 2 standard deviations.
Thus, given that t</span><span>he mean income per household in a certain state is $29,500 with a standard deviation of $5,750.
The values between which 95% of the incomes lies is obtained as follows:
![95\% \ of \ the \ data \ lies \ between \ \mu\pm2\sigma=\$29,500\pm2(\$5,750) \\ \\ =\$29,500-\$11,500 \ and \ \$29,500+\$11,500=\bold{\$18,000 \ and \ \$41,000}](https://tex.z-dn.net/?f=95%5C%25%20%5C%20of%20%5C%20the%20%5C%20data%20%5C%20lies%20%5C%20between%20%5C%20%5Cmu%5Cpm2%5Csigma%3D%5C%2429%2C500%5Cpm2%28%5C%245%2C750%29%20%5C%5C%20%20%5C%5C%20%3D%5C%2429%2C500-%5C%2411%2C500%20%5C%20and%20%5C%20%5C%2429%2C500%2B%5C%2411%2C500%3D%5Cbold%7B%5C%2418%2C000%20%5C%20and%20%5C%20%5C%2441%2C000%7D)
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