The heights of adult men in america are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches. the heights of adult women in america are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.53 inches. a) if a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? z = 2.16 b) what percentage of men are shorter than 6 feet 3 inches? round to nearest tenth of a percent.

z-score is 2.16

98.46% of men are SHORTER than 6 feet 3 inches

Step-by-step explanation:

The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches.

Mean =

Standard deviation =

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)

x = 6 feet 3 inches

1 feet =12 inches

6 feet = 12*6 = 72 inches

So, x = 6 feet 3 inches  = 72+3=75 inches

Formula :

So, his z-score is 2.16

No to find percentage of men are SHORTER than 6 feet 3 inches

We are supposed to find P(x< 6 feet 3 inches)

z-score is 2.16

Refer the z table

P(x< 6 feet 3 inches) =0.9846

So, 98.46% of men are SHORTER than 6 feet 3 inches

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step-by-step explanation:

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6