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kati45 [8]
3 years ago
12

PLEASE HELP ME ASAP!!!

Mathematics
1 answer:
statuscvo [17]3 years ago
6 0
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}

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}



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)

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}



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)

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}

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}</span>
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