Respuesta :
Answer:
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.
Answer: 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
