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Answer:
0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Also, important to remember that the standard deviation is the square root of the variance.
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 random variable X, with mean and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the sample mean would be greater than 70.5 kilograms?
This is 1 subtracted by the pvalue of Z when X = 70.5. So
By the Central limit theorem
has a pvalue of 0.7764
1 - 0.7764 = 0.2236
0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.
<h2>Answer:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>
(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is </u></em>