Question

Multiple question needing guidance please.

1. A distribution of values is normal with a mean of 173.2 and a
standard deviation of 39.

Find the probability that a randomly selected value is between
200.5 and 298.

*P*(200.5 < *X* < 298) =_______________

2. A distribution of values is normal with a mean of 229.7 and a
standard deviation of 83.5.

Find *P*_{32}, which is the score separating the
bottom 32% from the top 68%.

*P*_{32} = __________________

3. 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 5.9-in and a standard deviation
of 1-in. Due to financial constraints, the helmets will be designed
to fit all men except those with head breadths that are in the
smallest 2.4% or largest 2.4%.

What is the minimum head breadth that will fit the clientele?

min = ________________

What is the maximum head breadth that will fit the clientele?

max = __________________

Enter your answer as a number accurate to 1 decimal place. Answers
obtained using exact *z*-scores or *z*-scores rounded
to 3 decimal places are accepted.

Answer #1

1)

µ = 173.2

σ = 39

we need to calculate probability for ,

P ( 200.5 < X <
298 )

=P( (200.5-173.2)/39 < (X-µ)/σ < (298-173.2)/39 )

P ( 0.700 < Z <
3.200 )

= P ( Z < 3.200 ) - P ( Z
< 0.700 ) =
0.9993 - 0.7580 =
0.2413 (answer)

excel formula for probability from z score is
=NORMSDIST(Z)

2)

µ= 229.7

σ = 83.5

proportion= 0.32

Z value at 0.32 = -0.468
(excel formula =NORMSINV( 0.32 ) )

z=(x-µ)/σ

so, X=zσ+µ= -0.468 * 83.5
+ 229.7

X = 190.6 (answer)

3)

µ= 5.9

σ = 1

proportion= 0.024

Z value at 0.024 = -1.977
(excel formula =NORMSINV( 0.024 ) )

z=(x-µ)/σ

so, X=zσ+µ= -1.977 * 1
+ 5.9

X = 3.923 (answer)

**minimum head breadth that will fit the clientele = 3.9
in **

**---------------------------------**

µ= 5.9

σ = 1

proportion= 1-0.024 = 0.976

Z value at 0.976 = 1.977
(excel formula =NORMSINV( 0.976 ) )

z=(x-µ)/σ

so, X=zσ+µ= 1.977 * 1
+ 5.9

X = 7.877 (answer)

**maximum head breadth that will fit the clientele = 7.9
in**

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