Question

Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.. If 25 people are randomly selected, find the probability that their mean IQ score is less than 103. (a) .1587 (b) .8413 (c) 1.000 (d) .9938 23 Refer to question 19 above. If 100 people are randomly selected, find the probability that their mean IQ is greater than 103. (a) .8413 (b) 2.000 (c) .9772 (d) .0228 24 True or False. Because the total area under the normal standard distribution is equal to 1, there is a correspondence between area and probability. (a) True (b) False 25 True of False. A Z-score must be negative whenever it is located in the left half of the normal distribution. (a) True (b) False 26. The World Health Organization states that tobacco is the second leading cause of death in the world. Every year, a mean of 5 million people die of tobacco-related causes. Assume that the distribution is normal with µ = 5 (in millions) and ơ = 2 (in millions). Find the probability that more than 4 million people will die of tobacco-related causes in a particular year. (a) .3085 (b) .5000 (c) .6915 (d) -.5000 (e) .9772 27. Refer to question 26. What is the probability that the number of people who will die from tobacco-related deaths will fall between 3 million and 7 million in a particular year? (a) .1587 (b) .8413 (c) 1.000 (d) -1.000 (e) .6826 SHOW WORK

Answer 1

Solution :

Given that ,

mean = = 100

standard deviation = = 15

n = 25

= 10 0 and

= / n = 15 / 25 = 15 / 5 = 3

P( < 103) = P(( - ) / < (103 - 100) / 3)

= P(z < 1)

Using standard normal table,

= 0.8413

Probability = 0.8413

Option b)

(b)

mean = = 100

standard deviation = = 15

n = 100

= 100 and

= / n = 15 / 100 = 15 / 10 = 1.5

P( > 103) = 1 - P( < 03)

= 1 - P(( - ) / < (103 - 100) / 1.5)

= 1 - P(z < 2)

= 1 - 0.9772

= 0.0228

Probability = 0.0228

c)

True

Because the total area under the normal standard distribution is equal to 1,

there is a correspondence between area and probability .

d)

True

Z-score must be negative whenever it is located in the left half of the normal distribution .

e)

Given that ,

mean = = 5

standard deviation = = 2

P(x > 4) = 1 - P(x < 4)

= 1 - P((x - ) / < (4 - 5) / 2)

= 1 - P(z < -0.5)

= 1 - 0.3085   

= 0.6915

Probability = 0.6915

(e)

P(3 < x < 7) = P((3 - 5)/ 2) < (x - ) /  < (7 - 5) / 2) )

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413 - 0.1587

= 0.6826

Probability = 0.6826