Pleas help due to night
2 answers:
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Answer:
The students have to score 0.74 standard deviations above the mean to be publicly recognized.
Step-by-step explanation:
A random variable <em>X</em> is said to have a normal distribution with parameters <em>µ</em> (mean) and <em>σ</em>² (variance).
If , then
, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is,
.
The distribution of these <em>z</em>-scores is known as the standard normal distribution.
The <em>z</em>-score is a standardized form of the raw score, <em>X</em>. It is a numerical measurement of the relationship between a value (<em>X</em>) and the mean (<em>µ</em>) in terms of the standard deviation (<em>σ</em>). A <em>z</em>-score of -1 implies that the data value is 1 standard deviation below the mean. And a <em>z</em>-score of 1 implies that the data value is 1 standard deviation above the mean.
Let <em>X</em> be defined as the scores of students at the National Financial Capability Challenge Exam.
It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.
That is, P (X > x) = 0.23.
⇒
⇒
The value of <em>z</em> for this probability value is:
<em>z</em> = 0.74.
*Use a <em>z</em>-table.
Thus, the students have to score 0.74 standard deviations above the mean to be publicly recognized.
Answer:
The probability is
Step-by-step explanation:
From the question we are told that
The proportion that live with their parents is
The sample size is n = 125
Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution
i.e
Now the mean is evaluated as
Generally the proportion that are not staying with parents is
= >
The standard deviation is mathematically evaluated as
Given the n is large then we can use normal approximation to evaluate the probability as follows
Now applying continuity correction
Generally
So for the z - table