Answer:
option D
Step-by-step explanation:
Jerry solved this equation:
In the first step , she multiplied 3 inside the parenthesis
1.
In step 2, 3/4 is added on both sides
2.
In step 3, we take LCD 12
3.
In step 4, add the fractions
4.
Here, 3 or 3/1 are same.
In step 5, to remove 3/1 we multiply both sides by 1/3.
Instead of multiplying 1/3 , Jerry made an error by multiplying 3/1
In step 5, he should have multiplied both sides by 1/3
Answer:
a) By the Central Limit Theorem, it is approximately normal.
b) The standard error of the distribution of the sample mean is 1.8333.
c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.
d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours
e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours
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 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 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.
Mean of 36 hours and a standard deviation of 5.5 hours.
This means that
a. What can you say about the shape of the distribution of the sample mean?
By the Central Limit Theorem, it is approximately normal.
b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)
Sample of 9 means that . So
The standard error of the distribution of the sample mean is 1.8333.
c. What proportion of the samples will have a mean useful life of more than 38 hours?
This is 1 subtracted by the pvalue of Z when X = 38. So
By the Central Limit Theorem
has a pvalue of 0.8621
1 - 0.8621 = 0.1379
0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.
d. What proportion of the sample will have a mean useful life greater than 34.5 hours?
This is 1 subtracted by the pvalue of Z when X = 34.5. So
has a pvalue of 0.2061.
1 - 0.2061 = 0.7939
0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.
e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?
pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So
0.8621 - 0.2061 = 0.656
0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours
Answer: 0.9726
Step-by-step explanation:
Let x be the random variable that represents the distance the tires can run until they wear out.
Given : The top-selling Red and Voss tire is rated 50,000 miles, which means nothing. In fact, the distance the tires can run until they wear out is a normally distributed random variable with a 67,000 miles and a
5,200 miles.
Then , the probability that a tire wears out before 60,000 miles :
[using p-value table for z]
Hence, the probability that a tire wears out before 60,000 miles= 0.9726
Answer:
The test statistic value using the VLBW babies as group 1 is z=-2.76±0.01 and the P-value for the test (±0.0001) is 0.0030.
Step-by-step explanation:
This is a hypothesis test for the difference between proportions.
The claim is that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample 1 (VLBW group), of size n1=244 has a proportion of p1=0.77049.
The sample 2 (control group), of size n2=247 has a proportion of p2=0.79757.
The difference between proportions is (p1-p2)=-0.02708.
The pooled proportion, needed to calculate the standard error, is:
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the z-statistic as:
This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):
As the P-value (0.0030) is smaller than the significance level (0.05), the effect issignificant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.