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Mashcka [7]
3 years ago
6

At Stephanie's new job, she spent $6.50, $8, $7.25, $13.50, and $9.75 on lunch the first week. In the second week, Stephanie spe

nt a total of $10 less than the total of the first week. What is the decrease in the mean for the second week compared to the first week?
answer as a decimal rounded to the nearest hundredth.
Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0

Answer:

The mean for the second week is $2 less than the first and in percentage it is 22% less.

Step-by-step explanation:

The mean is given by the sum of all individual values divided by the number of values. For the first week the sum is:

sum1 = 6.5 + 8 + 7.25 + 13.5 + 9.75

sum1 = 45

Since she spent 10 less in the second week the sum is:

sum2 = sum1 - 10 = 45 - 10 = 35

The mean for each week is:

mean1 = sum1/5 = 45/5 = 9

mean2 = sum2/5 = 35/5 = 7

difference = mean1 - mean2 = 9-7 = 2

difference(%) = [2/9]*100 = 0.22*100 = 22%

The mean for the second week is $2 less than the first and in percentage it is 22% less.

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Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

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