## “Scores on an English test are normally distributed with a mean of 37.4 and a standard deviation of 7.9. Find the score that separates the t

Question

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## Answers ( )

Answer:Score that separates the top 59% from the bottom 41% is 35.6Step-by-step explanation:We are given that Scores on an English test are normally distributed with a mean of 37.4 and a standard deviation of 7.9, i.e.; = 37.4 and = 7.9

Now, the z score probability distribution is given by;

Z = ~ N(0,1)

The bottom 41% area is given by the critical z value of -0.2278 (from z% table)

So, P(Z < ) = 0.41

which means = -0.2278

X – 37.4 = -0.2278 * 7.9

X = 37.4 – 1.79962 = 35.6

Therefore, score of 35.6 separates the top 59% from the bottom 41%.Answer:The score that separates the top 59% from the bottom 41% is 35.6225

Step-by-step explanation:Problems of normally distributed samples are 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.

In this problem, we have that:Find the score that separates the top 59% from the bottom 41%”This is the value of X when Z has a pvalue of 0.41. So it is X when Z = -0.225.

The score that separates the top 59% from the bottom 41% is 35.6225