HW # 7 Area Under Normal Distribution Assuming that height data on a valid sampl

Question

HW # 7 Area Under Normal Distribution Assuming that height data on a valid sample of NNSemployees was collected and analyzed, producing the following: Average 5'8 Standard Deviation4" Calculate: a) Height range (centered around the mean ) that would indude 68% of the shipyard employees? b) Height of a doorway that will allow 90%of employees to walk under? C) Height of a doorway that will cause 3%of employees to hit their heads? d) Percent of employees who would hit their head on 5' 10" doorway? e) Percent of employees between 52" and 5'9"?

Explanation / Answer

Percentage of employees which will be above upper end of height range = 32/2 = 16%

Percentage of employees which will be below lower end of height range = 16%

We therefore first need to find :

Maximum height which will cover 84% ( i.e probability of 0.84 ) of people

Maximum height which will cover 16% ( i.e. probability 0.16 ) of people

Z value of probability 0.16 = NORMSINV ( 0.16) = - 0.9944

Therefore maximum height

= Average height + z value x Standard deviation of height

= 5 ft 8 inches + 0.9944 x 4 inches

= 5 ft 8 inches + 3.977 inches

= 11.977 inches

Minimum height

= Average height -Z value x Standard deviation of height

= 5 ft 8 inches – 0.994 x 4 inches

= 5 ft 8 inches – 3.977 inches

= 5 ft 4.023 inches

Height range that would include 68% of the shipyard employees will be between 5 ft 4.023 inches and 11.977 inches

Corresponding z value for probability 0.90 = NORMSINV ( 0.90 ) = 1.2815

Required height of the doorway

= Average height of doorway + z value x Standard deviation of height of doorway

= 5 ft 8 inches + 1.2815 x 4 inches

= 5 ft 8 inches + 5.126 inches

= 6 ft 1.126 inches

HEIGHT OF A DOORWAY THAT WILL ALLOW90% OF EMPLOYEES TO WALKUNER = 6 FT 1.126 INCHES

               Corresponding Z value for probability of 0.97 ( i.e. 97 % ) = NORMSINV ( 0.97 ) = 1.880

Required height of doorway

= Average height + z value x standard deviation of height

= 5 ft 8 inches + 1,88 x 4 inches

= 5 ft 8 inches + 7.52 inches

= 6 ft 3.52 inches

HEIGHT OF A DOORWAY THAT WILL CAUSE 3% OF THEIR EMPLOYEES TO HIT THEIR HEADS WILL BE 6 FT 3.52 INCHES

Therefore ,

Average height + Z1 x Standard deviation of height = 5 ft 10 inches

Or, 5 ft 8 inches + Z1 x 4 inches = 5 ft 10 inches

Or, Z1 X 4 inches = 2 inches

Or, Z1 = 0.5

Probability corresponding to Z = 0.5 will be 0.6915

Therefore , percentage of people ,that will walk under 5 ft 10 inches doorway = 69.15%

Hence, percentage of employees who will hit their head on 5 ft 10 inches doorway = 100% - 69.15% = 30.85%

PERCENTAGE OF EMPLOYEES WHO WILL HIT THEIR HEAD ON 5 FT 10 INCHES DOORWAY WILL BE 30.85%

Therefore,

Average height + Z1 x Standard deviation of height = 5 ft 9 inches

Or, 5 ft 8 inches + Z1 x 4 inches = 5 ft 9 inches

Or, Z1 x 4 inches = 1 inch

Or, Z1 = 0.25

Corresponding probability for Z= 0.25 as derived from standard normal distribution table 0.69871

Let z value corresponding to probability that employees height will be maximum 5 ft 2 inches is Z2

Therefore,

Average height + Z2 x Standard deviation of height = 5 ft 2 inches

Or, 5 ft 8 inches + Z2 x 4 inches = 5 ft 2 inches

Or, Z2 x 4 inches = - 6 inches

Or, Z2 = - 1.5

Corresponding probability for Z2 = - 1.5 will be 0.0668

Probability that employees will be between 5 ft 2 inches and 5 ft 9 inches = 0.69871 – 0.0668 = 0.6319 ( 0r, 63,19%)

PERCENTAGE OF EMPLOYEES BETWEEN 5 FT 2 INCHES AND 5 FT 9 INCHES = 63.19 %

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