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Komok [63]
3 years ago
15

uppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.2 and a standard deviation of 1.49. Us

ing the empirical rule, what percentage of American women have shoe sizes that are less than 12.67
Mathematics
1 answer:
liubo4ka [24]3 years ago
6 0

Answer:

99.85% of American women have shoe sizes that are less than 12.67

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 8.2

Standard deviation = 1.49

Using the empirical rule, what percentage of American women have shoe sizes that are less than 12.67

12.67 = 8.2 + 3*1.49

12.67 is 3 standard deviations above the mean.

Since the normal distribution is symmetric, 50% of the measures are below the mean and 50% are above. Of those 50% above, 99.7% are between the mean and 12.67. So

0.5 + 0.997*0.5 = 0.9985

99.85% of American women have shoe sizes that are less than 12.67

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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
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Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

The degrees of freedom are given by:

df=n-1=93-1=92  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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