Question

Hint: For z-scores with a magnitude greater than 3.8, we know from the z-table that the probability of a random case or sample mean being further from the population mean is less than .0001. This is true if z is 3.9, -4, or 24. Basically, if a case is that much of an outlier, it’s really unlikely

1. In Wisconsin in 2014, there was a mean of 1128.8 injury-related emergency room visits per day (population mean), with a standard deviation of 349.3 (population standard deviation). a) In a random sample of 81 days, what is the probability that the sample mean for number of injury-related emergency room visits in Wisconsin is lower than 1000? b) In a random sample of 49 days, what is the probability that the sample mean for this variable is between 1,100 and 1,200?

2. Average beer consumption per capita in America is 78.4 liters per year, with a standard deviation of 37.6 liters. a) What is the probability that a random sample of 90 Americans produces a mean for beer consumption between 70 and 75 liters per year? b) What is the probability that a random sample of 10 Americans would drink on average more than 100 liters in a year?

3. Last season, the Milwaukee Bucks’ Giannis Antetokounmpo scored a mean of 26.9 points per game, with a standard deviation of 6.2 points. What is the probability that, in a random sample of 9 games from last season, Giannis would average over 30 points per game?

Answer 1

(1)

(a)

= 1128.8

= 349.3

n = 81

SE = /

= 349.3/

= 38.8111

To find P(<1000):

Z = (1000 - 1128.8)/38.8111

= - 3.3186

Table of Area Under Standard Normal Curve gives area = 0.4995

So

P(<1000) = 0.5 - 0.4995 = 0.0005

So,

Answer is:

0.0005

(b)

= 1128.8

= 349.3

n = 49

SE = /

= 349.3/

= 49.9

To find P(1100< <1200):

Case 1: For X from 1100 to mid value:

Z = (1100 - 1128.8)/49.9

= - 0.5772

Table of Area Under Standard Normal Curve gives area = 0.2190

Case 2: For X from mid value to 1200:

Z = (1200 - 1128.8)/49.9

= 1.4269

Table of Area Under Standard Normal Curve gives area = 0.4236

So

P(1100<<1200) = 0.2190 + 0.4236 = 0.6426

So,

Answer is:

0.6426

(2)

(a)

= 78.4

= 37.6

n = 90

SE = /

= 37.6/

= 3.9634

To find P(70< <75):

Case 1: For X from 70 to mid value:

Z = (70 - 78.4)/3.9634

= - 2.1194

Table of Area Under Standard Normal Curve gives area = 0.4830

Case 2: For X from 75 to mid value:

Z = (75 - 78.4)/3.9634

= - 0.8578

Table of Area Under Standard Normal Curve gives area = 0.3051

So

P(70<<75) = 0.4830 - 0.3051 = 0.1779

So,

Answer is:

0.1779

(b)

= 78.4

= 37.6

n = 10

SE = /

= 37.6/

= 11.8902

To find P( >100):

Z = (100 - 78.4)/11.8902

= 1.8166

Table of Area Under Standard Normal Curve gives area = 0.4656

So

P(>100) = 0.5 - 0.4656 = 0.0344

So,

Answer is:

0.0344

3.

= 26.9

= 6.2

n = 9

SE = /

= 6.2/

= 2.0667

To find P( >30):

Z = (30 - 26.9)/2.0667

= 1.50

Table of Area Under Standard Normal Curve gives area = 0.4332

So

P(>30) = 0.5 - 0.4332 = 0.0668

So,

Answer is:

0.0668