QuestionAcme Company widget weights are normally distributed: mean $56$ oz, SD $7$ oz. Use the Empirical Rule to find:
a) Range for $68\%$ weights. b) Percentage between $56$ and $70$ oz. c) Percentage between $35$ and $77$ oz.

Studdy Solution

STEP 1

Assumptions1. The distribution of widget weights is bell-shaped, which suggests a normal distribution. . The mean (average) widget weight is56 ounces.
3. The standard deviation of the widget weights is7 ounces.
4. We will use the Empirical Rule (also known as the68-95-99.7 rule) which states that for a normal distribution, approximately68% of the data falls within one standard deviation of the mean,95% falls within two standard deviations, and99.7% falls within three standard deviations.

STEP 2

For part a), the Empirical Rule tells us that68% of the widget weights lie between one standard deviation below the mean and one standard deviation above the mean.
$68\% = Mean - Standard\, Deviation \,to\, Mean + Standard\, Deviation$

STEP 3

Now, plug in the given values for the mean and standard deviation to calculate the range.
$68\% =56 -7 \,to\,56 +7$

STEP 4

Calculate the range.
$68\% =49 \,to\,63$ So,68% of the widget weights lie between49 and63 ounces.

STEP 5

For part b), we need to find the percentage of widget weights that lie between the mean and one standard deviation above the mean.
Since the Empirical Rule tells us that68% of the data falls within one standard deviation of the mean, and this is symmetrical, we can say that half of this percentage (34%) lies between the mean and one standard deviation above the mean.
$Percentage =68\% /2$

STEP 6

Calculate the percentage.
$Percentage =68\% /2 =34\%$ So,34% of the widget weights lie between56 and70 ounces.

STEP 7

For part c), we need to find the percentage of widget weights that lie between35 and77 ounces.
First, calculate how many standard deviations35 and77 are from the mean.
$Standard\, Deviations\, Below\, Mean = (Mean - Lower\, Limit) / Standard\, Deviation$ $Standard\, Deviations\, Above\, Mean = (Upper\, Limit - Mean) / Standard\, Deviation$

STEP 8

Plug in the given values for the mean, standard deviation, lower limit, and upper limit to calculate the number of standard deviations.
$Standard\, Deviations\, Below\, Mean = (56 -35) /7 =3$ $Standard\, Deviations\, Above\, Mean = (77 -56) /7 =3$

STEP 9

According to the Empirical Rule, approximately99.7% of the data falls within three standard deviations of the mean.
So, approximately99.7% of the widget weights lie between35 and77 ounces.

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QuestionAcme Company widget weights are normally distributed: mean $56$ oz, SD $7$ oz. Use the Empirical Rule to find:
a) Range for $68\%$ weights. b) Percentage between $56$ and $70$ oz. c) Percentage between $35$ and $77$ oz.

Studdy Solution

STEP 1

STEP 2

For part a), the Empirical Rule tells us that68% of the widget weights lie between one standard deviation below the mean and one standard deviation above the mean.
$68\% = Mean - Standard\, Deviation \,to\, Mean + Standard\, Deviation$

STEP 3

Now, plug in the given values for the mean and standard deviation to calculate the range.
$68\% =56 -7 \,to\,56 +7$

STEP 4

Calculate the range.
$68\% =49 \,to\,63$ So,68% of the widget weights lie between49 and63 ounces.

STEP 5

STEP 6

Calculate the percentage.
$Percentage =68\% /2 =34\%$ So,34% of the widget weights lie between56 and70 ounces.

STEP 7

STEP 8

Plug in the given values for the mean, standard deviation, lower limit, and upper limit to calculate the number of standard deviations.
$Standard\, Deviations\, Below\, Mean = (56 -35) /7 =3$ $Standard\, Deviations\, Above\, Mean = (77 -56) /7 =3$

STEP 9

According to the Empirical Rule, approximately99.7% of the data falls within three standard deviations of the mean.
So, approximately99.7% of the widget weights lie between35 and77 ounces.

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QuestionAcme Company widget weights are normally distributed: mean 565656 oz, SD 777 oz. Use the Empirical Rule to find:a) Range for 68%68\%68% weights. b) Percentage between 565656 and 707070 oz. c) Percentage between 353535 and 777777 oz.

Studdy Solution

STEP 1

STEP 2

STEP 3

Now, plug in the given values for the mean and standard deviation to calculate the range.68%=56−7 to 56+768\% =56 -7 \,to\,56 +768%=56−7to56+7

STEP 4

Calculate the range.68%=49 to 6368\% =49 \,to\,6368%=49to63So,68% of the widget weights lie between49 and63 ounces.

STEP 5

STEP 6

Calculate the percentage.Percentage=68%/2=34%Percentage =68\% /2 =34\%Percentage=68%/2=34%So,34% of the widget weights lie between56 and70 ounces.

STEP 7

STEP 8

STEP 9

According to the Empirical Rule, approximately99.7% of the data falls within three standard deviations of the mean.So, approximately99.7% of the widget weights lie between35 and77 ounces.

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QuestionAcme Company widget weights are normally distributed: mean $56$ oz, SD $7$ oz. Use the Empirical Rule to find:
a) Range for $68\%$ weights. b) Percentage between $56$ and $70$ oz. c) Percentage between $35$ and $77$ oz.

Now, plug in the given values for the mean and standard deviation to calculate the range.
$68\% =56 -7 \,to\,56 +7$

Calculate the range.
$68\% =49 \,to\,63$ So,68% of the widget weights lie between49 and63 ounces.

Calculate the percentage.
$Percentage =68\% /2 =34\%$ So,34% of the widget weights lie between56 and70 ounces.

According to the Empirical Rule, approximately99.7% of the data falls within three standard deviations of the mean.
So, approximately99.7% of the widget weights lie between35 and77 ounces.