Plzzzz help!!!
1 answer:
You might be interested in
Answer:
One can be 99% confident the true mean shell length lies within the above interval.
The population has a relative frequency distribution that is approximately normal.
Step-by-step explanation:
We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.
The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;
P.Q. = ~
where, = sample mean shell length = 51.3 cm
s = sample standard deviation = 6.6 cm
n = sample of turtles = 6
= true mean shell length
Now, the 99% confidence interval for =
Here, = 1% so
= 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.
<u>So, 99% confidence interval for</u> =
= [51.3 - 10.864 , 51.3 + 10.864]
= [40.44 cm, 62.16 cm]
The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].
The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;
C. The population has a relative frequency distribution that is approximately normal.
This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.
Answer:
a) Mean blood pressure for people in China.
b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.
c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.
d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.
e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg
f) 157.44mmHg
Step-by-step explanation:
When the distribution is normal, we use 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.
If X is two standard deviations from the mean or more, it is considered unusual.
In this question:
a.) State the random variable.
Mean blood pressure for people in China.
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
This is 1 subtracted by the pvalue of Z when X = 135.
has a pvalue of 0.6179
1 - 0.6179 = 0.3821
38.21% probability that a person in China has blood pressure of 135 mmHg or more.
c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
This is the pvalue of Z when X = 141.
has a pvalue of 0.7140
71.30% probability that a person in China has blood pressure of 141 mmHg or less.
d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So
X = 125
has a pvalue of 0.4483
X = 120
has a pvalue of 0.3632
0.4483 - 0.3632 = 0.0851
8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
From b), when X = 135, Z = 0.3
Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.
f.) What blood pressure do 90% of all people in China have less than?
This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then
X = 120
So
157.44mmHg