Dr. Spike is a big fan of Bojangles and is particularly interested in the popularity of its celebrated wings dinner. Starting fr
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Answer:
Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)
Step-by-step explanation:
Given: Probability of a children under age 5 has food allergies = 7%
=
To find : Probability of atleast one of 12 student has food allergies
Probability of a chindren under age 5 does not have food allergies =
⇒ prob =
now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.
But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)
Probability of 0 student having allergy out of 12 student =
Therefore, Probability of atleast one of 12 student has food allergies
=
=
≈ 0.58 ( approx)
Answer:
26.11% of the test scores during the past year exceeded 83.
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 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.
In this problem, we have that:
It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that .
Approximately what percentatge of the test scores during the past year exceeded 83?
This is 1 subtracted by the pvalue of Z when . So:
has a pvalue of 0.7389.
This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.
Answer:
The co variance of the midterm and final exam scores is 58.76.
Step-by-step explanation:
The formula to compute the sample co variance is:
The values are computed in the table below.
Compute the co variance as follows:
Thus, the co variance of the midterm and final exam scores is 58.76.
