Question

Countdown: Days:-11 Time:-07:50:40 checkpoint #2: Examining Distributions cont'd Schedule: Available now. Due 02/04/2020 11:59 PM EST Page 1 Question 1 Select one answer. 10 points The distribution of the IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 15. According to the standard deviation rule, only 0.15% of people have an IQ over what score? 55 100 115 130 145 Question 2 Select one answer. 10 points The distribution of the amount of money spent by first-time gamblers at a major casino in Las Vegas is approximately normal in shape with a mean of $600 and a standard deviation of $120. According to the standard deviation rule, almost 84% of gamblers spent more than what amount of money at this casino? $360 $480 $600 $720 $840 Question 3 Select one answer. 10 points The histogram below displays the distribution of 50 ages at death due to trauma (unnatural accidents and homicides) that were observed in a certain hospital during a week.

_u2_summarizing_data/_m1_examining_distributions/webcontent/assessment1q2.gif Which of the following are the appropriate numerical measures to describe the center and spread of the above distribution? The IQR and the standard deviation The mean and the standard deviation The median and the IQR The mean and the median

Answer 1

Question 1

Answer: 145

Explanation:

Let only 0.15% of people have an IQ above score X,

hence, % of people below X = 99.85%.

Since the score is normally distributed with a mean of 100 and a standard deviation of 15

The z score is obtained from the standard normal distribution table for 99.85% area of the standard normal curve(in excel use function =NORM.S.INV(0.9985))

Question 2

Answer: $480

Explanation:

Let only 84% of people have an IQ above score X,

hence, % of people below X = 16%.

Since the score is normally distributed with a mean of 600 and a standard deviation of 120

The z score is obtained from the standard normal distribution table for 16% area of the standard normal curve(in excel use function =NORM.S.INV(0.16))

(Image is not uploaded for question 3)