Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.6 feet and a standard deviation of 0.2 feet. A sample of 53 men’s step lengths is taken. Step 1 of 2 : Find the probability that an individual man’s step length is less than 2.4 feet. Round your answer to 4 decimal places, if necessary.

P[ X < 2,4 ]  = 0,1587   or   15,87 %

Step-by-step explanation:

Normal Distribution  N ( 2,6 , 0,2 )

mean = μ₀ = 2,6 ft  and

standard dviation is

σ = 0,2 ft

Samle size 53    n  = 53

2,4 - 2,6 = 0,2

That means the difference between 2,4 and the mean of the sample ( 2,6) is equal to one standar deviation.  The empircal rule establishes that the interval [  μ₀ ± σ]  contans 68,3 % of all values, by symmetry

μ₀ - σ = 68,3/2

μ₀ - σ = 34,15

Then the half of the bell shape curve is 0,5 and  from 0,2 up to 0 ( the normalized μ₀ ) is 34,15 then dfference between

0,5 - 0,3415   =  0,1585  or   15,85% is the probability

Other procedure is:

P[ X < 2,4 ]  = [X  -  μ₀ ] / σ

P[ X < 2,4 ]  = 2,4 - 2,6 / 0,2

P[ X < 2,4 ]  = - 0,2/0,2

P[ X < 2,4 ]  = - 1

And fom z - table we find for z (score) = - 1

P[ X < 2,4 ]  = 0,1587

0.0668

Step-by-step explanation:

z=x−μσ=2.3−2.60.2=−1.50

Using the normal distribution tables, we find that the area under the standard normal curve to the left of z=−1.50 is approximately 0.0668. Thus, the probability that an individual man’s step length is less than 2.3 feet is approximately 0.0668.

m∠a is 90°

m∠b is 40°