5) How many thousands are there in 2 158 621?
2 answers:
You might be interested in
Answer:
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
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
Probability that no less than 92 out of 159 students will pass their college placement exams.
No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So
has a pvalue of 0.7157
1 - 0.7157 = 0.2843
0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.