Suppose that scores on a particular test are normally distributed with a mean of 140 and a standard

4 answers
Question:

Suppose that scores on a particular test are normally distributed with a mean of 140 and a standard deviation of 20. What is the minimum score needed to be in the top 5% of the scores on the test? Carry your intermediate computations to at least four decimal places, and round your answer to at least one decimal place.

Answers

The minimum score needed to be in the top 5% of the scores on the test is 172.9.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

In this problem, we have that:

[tex]\mu = 140, \sigma = 20[/tex]

What is the minimum score needed to be in the top 5% of the scores on the test?

The 100-5 = 95th percentile, which is the value of X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.645 = \frac{X - 140}{20}[/tex]

[tex]X - 140 = 20*1.645[/tex]

[tex]X = 172.9[/tex]

The minimum score needed to be in the top 5% of the scores on the test is 172.9.

the minimum score needed to be in the top 5% of the scores on the test is 173

Step-by-step explanation:

Suppose that scores on the particular test are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores on the test.

µ = mean score

σ = standard deviation

From the information given,

µ = 140

σ = 20

The probability value for the minimum score needed to be in the top 5% of the scores on the test would be (1 - 5/100) = (1 - 0.05) = 0.95

Looking at the normal distribution table, the z score corresponding to the probability value is 1.65

Therefore,

1.65 = (x - 140)/20

Cross multiplying by 20, it becomes

1.65 × 20 = x - 140

33 = x - 140

x = 33 + 140

x = 173

answer: 59

step-by-step explanation:

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒[tex]y=\frac{+1}{+1}x+0[/tex]

or   [tex]y=\frac{-1}{-1}x+0[/tex]

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

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