Help please. Will put brainliest
1 answer:
Answer:
- 17,750
- 1737
Step-by-step explanation:
A suitable table or calculator is needed.
One standard deviation from the mean includes 68.27% of the total, so the number of bottles in the range 20 ± 0.16 ounces will be ...
0.6827·26,000 = 17,750 . . . . . within 20 ± 0.16
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The number below 1.5 standard deviations below the mean is about 6.68%, so for the given sample size is expected to be ...
0.66799·26,000 = 1737 . . . . . below 19.76
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<em>Comment on the first number</em>
The "empirical rule" tells you that 68% of the population is within 1 standard deviation (0.16 ounces) of the mean. When the number involved is expected to be expressed to 5 significant digits, your probability value needs better accuracy than that. To 6 digits, the value is 0.682689, which gives the same "rounded to the nearest integer" value as the one shown above.
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Answer:
The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution should be used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 28 - 1 = 27
80% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.314
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 34.6 - 0.646 = $33.954
The upper end of the interval is the sample mean added to M. So it is 34.6 + 0.646 = $35.246
The 80% confidence level estimate for how much a typical parent would spend on their child's birthday gift is between $33.954 and $35.246.