Answer :
Answer:
0.964 is the probability that their mean shoulder breadth is less than 18.5 inch.
Step-by-step explanation:
Given:
Mean, μ = 18.2 inch
Standard Deviation, σ = 1.0 inch
n = 36
We are given that the distribution of shoulder breadths is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
a) P( mean shoulder breadth is less than 18.5 inch)
P(x < 18.5)
[tex]P( x < 18.5) = P( z < \displaystyle\frac{18.5-18.2}{\frac{1}{\sqrt{36}}}) = P(z < 1.8)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 18.5) =0.964= 96.4\%[/tex]
Thus, 0.964 is the probability that their mean shoulder breadth is less than 18.5 inch.
Yes, the result suggest that money can be saved by making smaller manholes with a diameter of 18.5 inch since 96.4% of the man holes have their mean shoulder breadth less than 18.5 inch.