The scatter plot shows how many bugs, with and without wings, were found by students on a field trip. How many bugs with wings d
id students Q, F, and G find?
A) 37
B) 44
C) 34
D) 27
A) 37
B) 44
C) 34
D) 27
2 answers:
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Answer: Please, see the attached file.
Thanks.
Solution:
We can graph using the intercepts:
(1) y-intercept, when x=0
y=4x-3→y=4(0)-3→y=0-3→y=-3
Point=(x,y)=(0,-3)
(2) x-intercept, when y=0
y=4x-3→0=4x-3
Solving for x: Adding 3 both sides of the equation:
0+3=4x-3+3
3=4x
Dividing both sides of the equation by 4:
3/4=4x/4
3/4=x
x=3/4=0.75
Point=(x,y)=(3/4,0)=(0.75,0)
With these pair of points (0, -3) and (0.75,0) we can draw the right line
Answer:
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
Step-by-step explanation:
To solve these questions, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes 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
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.
This means that
Sample of 10:
This means that
Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
This is 1 subtracted by the p-value of Z when X = 40. So
By the Central Limit Theorem
has a p-value of 0.8997
1 - 0.8997 = 0.1003
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
This is the p-value of Z when X = 260. So
has a p-value of 0.3539.
0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?
Now , so:
has a p-value of 0.0465.
0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?
Now , so:
has a p-value of 0.0040.
0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?
Sample of size 15 means that . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.
X = 276
has a p-value of 0.9922.
X = 256
has a p-value of 0.0078.
0.9922 - 0.0078 = 0.9844
0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.