In a recent survey of high school students, it was found that the average amount of money spent on entertainment each week was n
ormally distributed with a mean of $52.30 and a standard deviation of $18.23. Assuming these values are representative of all high school students, what is the probability that for a sample of 25, the average amount spent by each student exceeds $60?
A) 0.3372
B) 0.0174
C) 0.1628
D) 0.4826
A) 0.3372
B) 0.0174
C) 0.1628
D) 0.4826
2 answers:
Answer: A
Step-by-step explanation:
To solve this problem, you will first need to calculate the Z value of the variable of interest and then use a Z distribution table to look up the probability. To calculate the Z score, use the following formula.
See attachment
x = the value that is being standardized
m = the mean of the distribution
s = standard deviation of the distribution
Z>$60-$52.30/$18.23
Z>$7.7/$18.23
Z=>4224
then you check the Z value on the standard normal distribution table =.6628
1.0000-0.6628=0.3372
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Answer:
1. 13 or -13
2. -5 < y < -3
3. 6 or -6
4. 1/8 or -1/8
Step-by-step explanation:
Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.
The Absolute Value term is |x|
For the Negative case we'll use -(x)
For the Positive case we'll use (x)
Step 3 :
Solve the Negative Case
-(x) = 13
Multiply
-x = 13
Multiply both sides by (-1)
x = -13
Which is the solution for the Negative Case
Step 4 :
Solve the Positive Case
(x) = 13
Which is the solution for the Positive Case
Step 5 :
Wrap up the solution
x=-13
x=13
But for the case of question (2) its a different pattern..
Since this is a "less than" absolute-value inequality, my first step is to clear the absolute value according to the "less than" pattern. Then I'll solve the linear inequality.
| y + 4 | < 1
–1 < y + 4 < 1
This is the pattern for "less than". Continuing, I'll subtract 3 from all three "sides" of the inequality:
–1 – 4 < y + 4 - 4 < 1 – 4
–5 < y < -3
The solution to the original absolute-value inequality, | y + 4 | < 1 , is the interval: