For a certain population of sea turtles, 18 percent are longer than 6.5 feet. A random sample of 90 sea turtles will be selected
1 answer:
Answer:
The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405
Step-by-step explanation:
Central Limit Theorem:
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
18 percent are longer than 6.5 feet.
This means that
A random sample of 90 sea turtles
This means that
What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90
The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405
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Answer:
35.2% probability that the sample mean will be 246 pages or more
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:
What the probability that the sample mean will be 246 pages or more?
This is 1 subtracted by the pvalue of Z when X = 246. So
By the Central Limit Theorem
has a pvalue of 0.6480.
1 - 0.6480 = 0.3520
35.2% probability that the sample mean will be 246 pages or more
Answer:
a) P(X<50)=0.9827
b) P(X>47)=0.4321
c) P(-1.5<z<1.5)=0.8664
Step-by-step explanation:
We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.
a) This means we have to calculate P(x<50).
We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:
b) We have to calculatee P(x>47).
We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:
c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5
So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):