John has a 6 inch cuck gwen has 4 inch deep_______ how much of johns cuck is sticking out?
2 answers:
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Answer:
6.18% probability that more than 150 first-year students attend this college.
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
For each item selected, there are only two possible outcomes. Either it is defective, or it is not. This means that we use concepts of the binomial probability distribution to solve this problem.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
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:
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.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
Approximate the probability that more than 150 first-year students attend this college.
This is 1 subtracted by the pvalue of Z when X = 150. So
has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% probability that more than 150 first-year students attend this college.
Distance of each track are:
D₁ = 428.5 yd
D₂ = 436.35 yd
D₃ = 444.20 yd
D₄ = 452.05 yd
D₅ = 459.91 yd
D₆ = 467.76 yd
D₇ = 475.61 yd
D₈ = 483.47 yd
<u>Explanation:</u>
Given:
Track is divided into 8 lanes.
The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.
Thus,
Starting from the innermost edge with a diameter of 60yd.
Each lane is 10/8 = 1.25yd
So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.
So, the distances are:
D₁ = 240 + π (60) → 428.5yd
D₂ = 240 + π(60 + 2.5) → 436.35 yd
D₃ = 240 + π(60 + 5) → 444.20 yd
D₄ = 240 + π(60 + 7.5) → 452.05 yd
D₅ = 240 + π(60 + 10) → 459.91 yd
D₆ = 24 + π(60 + 12.5) → 467.76 yd
D₇ = 240 + π(60 + 15) → 475.61 yd
D₈ = 240 + π(60 + 17.5) → 483.47 yd