Respuesta :
Answer:
0.0668 = 6.68% probability that a student taking the test will score less than 400.
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 = 550, \sigma = 100[/tex]
Find the probability that a student taking the test will score less than 400.
This is the pvalue of Z when X = 400. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{400 - 550}{100}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
0.0668 = 6.68% probability that a student taking the test will score less than 400.
Answer:
Probability that a student taking the test will score less than 400 is 0.0668.
Step-by-step explanation:
We are given that scores on a college entrance exam are normally distributed with a mean of 550 and a standard deviation of 100.
Let, X = scores on a college entrance exam
X ~ N([tex]\mu = 550, \sigma = 100^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean score
[tex]\sigma[/tex] = standard deviation
So, probability that a student taking the test will score less than 400 is given by = P(X < 400)
P(X < 400) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{400-550}{100}[/tex] ) = P(Z < -1.50) = 1 - P(Z [tex]\leq[/tex] 1.50)
= 1 - 0.9332 = 0.0668
Therefore, probability that a student taking the test will score less than 400 is 0.0668.
