Suppose that salaries for recent graduates of one university have a mean of $25,700 with a standard deviation of $1350. Using Ch
1 answer:
Answer:
75% of the data will reside in the range 23000 to 28400.
Step-by-step explanation :
To find the range of values :
We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,
Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.
Range is given by : mean +/- (z score) × (value of a standard deviation)
⇒ Range : 25700 +/- 2 × 1350
⇒ Range : (25700 - 2700) to (25700 + 2700)
Hence, 75% of the data will reside in the range 23000 to 28400.
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Answer:
The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.28.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5
The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7
The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.