Question

PLEASE HELP ME ASAP!!!

1. The average cost per ounce for glass cleaner is 7.7 cents with a standard deviation of 2.5 cents. What is the z-score of glass cleaner with a cost of 10.1 cents per ounce?

A. 0.96
B. 1.31
C. 1.94
D. 4.04
E. none of the above



2. A candy bar takes an average of 3.4 hours to move through an assembly line. If the standard deviation is 0.5 hours, what is the probability that the candy bar will take between 3 and 4 hours to move through the line?

A. .2119
B. .2295
C. .6731
D. .3270
E. .8849



3. The average noise level in a diner is 30 decibels with a standard deviation of decibels. 99% of the time, the noise level is below what value?

A. 16.04
B. 30.00
C. 36.00
D. 43.96
E. 48.00



4. The mean income per household in a certain state is $29,500 with a standard deviation of $5,750. The middle 95% of incomes is between what two values? The incomes are normally distributed.

A. $23, 750 and $35,250
B. $18,000 and $41,000
C. $29,500 and $41,000
D. $12,250 and $46,750
E. $12,325 and $81,550

Answer 1

Part 1:

The z-score of a value, a, given the mean, μ, and the standard deviation, σ, is given by


`z= (a-\mu)/(\sigma)`

Thus, given that the 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= (10.1-7.7)/(2.5) = (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 \ (b-\mu)/(\sigma) \right)-P\left(z\ \textless \ (a-\mu)/(\sigma) \right)`

Thus, given that a candy bar takes an average of 3.4 hours to move through an assembly line and that the standard deviation is 0.5 hours.

The probability 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 \ (4-3.4)/(0.5) \right)-P\left(z\ \textless \ (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 the 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 is the value for which the probability that the noice level is below that value is 0.99.

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 \ (a-\mu)/(\sigma) \right)`

Thus, given that the 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 \ (a-30)/(4) \right)=P(z\ \textless \ 2.33) \\ \\ \Rightarrow(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 the 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}`