Engineers must consider the breadths of male heads when
designing helmets. The company researchers have determined that the
population of potential clientele have head breadths that are
normally distributed with a mean of 5.6-in and a standard deviation
of 0.8-in.
In what range would you expect to find the middle 95% of most head
breadths?
Between .................. and ...................
If you were to draw samples of size 47 from this population, in
what range would you expect to find the middle 95% of most averages
for the breadths of male heads in the sample?
Between ......................... and
........................
Enter your answers as numbers. Your answers should be accurate to 2
decimal places.
µ = 5.6
σ = 0.8
proportion= 0.9500
proportion left 0.0500 is equally distributed both left and right side of normal curve
z value at 0.025 = ± 1.960 (excel formula =NORMSINV( 0.05 / 2 ) )
z = ( x - µ ) / σ
so, X = z σ + µ =
X1 = -1.960 * 0.8 + 5.6 = 4.03
X2 = 1.960 * 0.8 + 5.6 = 7.17
.......
µ = 5.6
σ = 0.8
n= 47
proportion= 0.95
proportion left 0.05 is equally distributed both left and right side of normal curve
z value at 0.025 = ± -1.96 (excel formula =NORMSINV( 0.05 / 2 ) )
Z value at 0.975 = 1.96
z = ( x - µ ) / (σ/√n)
so, X = z σ / √n + µ =
X1 = -1.96 * 0.8 / √ 47 + 5.6 = 5.3713
X2 = 1.96 * 0.8 / √ 47 + 5.6 = 5.8287
...........
Thanks
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