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Question 10: The distribution of the amount of money spent by students on textbooks in a...

Question 10:

The distribution of the amount of money spent by students on textbooks in a semester is approximately normal in shape with a mean of 494 and a standard deviation of 39.

According to the standard deviation rule, approximately 68% of the students spent between $_____ and $ ______ on textbooks in a semester.

Question 11:

The distribution of IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 16.

According to the standard deviation rule, _____ % of people have an IQ between 52 and 148. Do not round.

Question 12:

The distribution of IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 19.

According to the standard deviation rule, only ______ % of people have an IQ over 157.

Question 13:

The distribution of the amount of money spent by students on textbooks in a semester is approximately normal in shape with a mean of: μ= 429 and a standard deviation of: σ= 23.

According to the standard deviation rule, almost 16% of the students spent more than what amount of money on textbooks in a semester?

Solutions

Expert Solution

Solution(10)

Given in the question

Mean = 494

Standard deviation = 39

According to empirical rule 68% of the data is +/- standard deviation from the mean

Lower value = (494-39) = 455

Upper value = (494+39) = 533

So we can say that 68% of students spent between $455 and $533

Solution(11)

Given in the question

Mean = 100

Standard deviation = 16

We need to calculate P(52<X<148) = P(X<148) - P(X<52)

Z = (52-100)/16 = -3

Z = (148-100)/16 = 3

From Z table we found p value

P(52<X<148) = 0.9987- 0.0013 = 0.9974

So there is 99.74% of people have an IQ between 52 and 148.

Solution(12)

Given in the question

Mean = 100

Standard deviation = 19

We need to calculate P(X>157) = 1- P(X<=157)

Z = (157-100)/19 = 3

From Z table we found p-value

P(X<157) = 1-0.9987 =0.0013

Solution(13)

Given in the question

Mean = 429

Standard deviation = 23

P-value = 0.84

So Z score from Z table = 1

So 1 =( X-429)/23

X = 429 +23 = 452


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