Given that a>0 compare: a+20/40 and a+10/20
1 answer:
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Answer:
a) 5.37% probability that an individual distance is greater than 210.9 cm
b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.
c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
a. Find the probability that an individual distance is greater than 210.9 cm
This is 1 subtracted by the pvalue of Z when X = 210.9. So
has a pvalue of 0.9463.
1 - 0.9463 = 0.0537
5.37% probability that an individual distance is greater than 210.9 cm.
b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.
Now
This probability is 1 subtracted by the pvalue of Z when X = 196. Then
By the Central Limit Theorem
has a pvalue of 0.2420.
1 - 0.2420 = 0.7580
75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.
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Answer:
- 20 months
- $5500
Step-by-step explanation:
There are a couple of ways to work a problem like this. You have probably been taught to write equations for each of the payment amounts as a function of time, then equate those values to solve for the time that makes them equal.
at dealer 1, the total amount paid (y) will be a function of months (x):
y = 2500 +150x
at dealer 2, the corresponding equation is ...
y = 3000 +125x
These are equal when ...
y = y
2500 +150x = 3000 +125x
25x = 500 . . . . . . . . . subtract 125x +2500 from both sides
x = 500/25 = 20
The total paid will be the same after 20 months.
That amount is ...
y = 2500 +150(20) = 5500
$5500 will be paid to either dealer after 20 months.
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The other way to work the problem is to "cut to the chase". The difference in down payment is made up at the rate of difference in monthly payments. So The number of monthly payments (x) required to equal the difference in down payments is ...
25x = 500 . . . . . . . . . you may recognize this equation from above
x = 500/25 = 20
