The time required for one wing beat is 5 milliseconds.
<em><u>Explanation</u></em>
Lets assume, the time required for one wing beat is seconds
So, the time required for 200 wing beat seconds
Given that, the honeybee flaps its wings 200 times per 1 second. So...
So, the time required for one wing beat is 5 milliseconds.
Answer:
0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.
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 problem, we have that:
Find the probability that the average score of the 64 golfers exceeded 76.
This is 1 subtracted by the pvalue of Z when X = 64.
By the Central Limit Theorem
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.
