Answer:
Step-by-step explanation:
We want to determine a 95% confidence interval for the true population mean textbook weight.
Number of sample, n = 40
Mean, u = 52 ounces
Standard deviation, s = 5.4 ounces
For a confidence level of 95%, the corresponding z value is 1.96.
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
52 ± 1.96 × 5.4/√40
= 52 ± 1.96 × 0.85
= 52 ± 1.666
The lower end of the confidence interval is 52 - 1.666 = 50.33 ounces
The upper end of the confidence interval is 52 + 1.666 = 53.67 ounces
Therefore, with 95% confidence interval, the true population mean textbook weight is between 50.33 ounces and 53.67 ounces
Answer:
63 ride the bus
Step-by-step explanation:
of means multiply
25% * 252
Change to decimal form
.25* 252
63
Answer:
=
0r
= 2.33
Step-by-step explanation:
=
0r
= 2.33
Answer:
The 90th percentile on this distribution is 12.57 organisms per liter.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This is the percentile that the score is in the distribution.
In this problem, we have that:
Water samples from a particular site demonstrate a mean coliform level of 10 organisms per liter with a standard deviation of 2, so .
What is the 90th percentile on this distribution?
Z has a pvalue of 0.90 when it is between and , so we use
The 90th percentile on this distribution is 12.57 organisms per liter.
Answer:
y=1
Step-by-step explanation:
4(1+2y)=12y
4*1 + 4*2y=12y
4 + 8y=12y
4=12y - 8y
4=4y
4/4=y
1=y