Respuesta :
Answer:
The percentage of adults in the town whose heights are between 60.1 and 74.1 inches
P( 60.1≤X≤74.1) = 2.26 %
Step-by-step explanation:
Step(i):-
Given that the heights of the adults in one town have a mean of 67.1 inches
Mean of the Population = 67.1 inches
The standard deviation of the Population = 3.5inches
Let X₁ =60.1
[tex]Z = \frac{x-mean}{S.D} = \frac{60.1-67.1}{3.5} = -2[/tex]
Let X₂ = 74.1
[tex]Z = \frac{x-mean}{S.D} = \frac{60.1-74.1}{3.5} = -4[/tex]
Step(ii):-
The probability of adults in the town whose heights are between 60.1 and 74.1 inches
P( X₁ ≤X≤X₂) = P(Z₁≤Z≤Z₂)
= P( -2≤ Z≤-4)
= |A(-4) -A(-2)|
= A(4) - A(2)| (∵ A(-Z) =A(Z))
= 0.4998 - 0.4772
= 0.0226
Final answer:-
The probability of adults in the town whose heights are between 60.1 and 74.1 inches
P( X₁ ≤X≤X₂) = 0.0226
The percentage of adults in the town whose heights are between 60.1 and 74.1 inches
P( 60.1≤X≤74.1) = 2.26%
95.44% of adults in the town have heights between 60.1 and 74.1 inches
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 67.1, σ = 3.5.
For x = 60.1:
[tex]z=\frac{60.1-67.1}{3.5}=-2[/tex]
For x = 74.1:
[tex]z=\frac{74.1-67.1}{3.5}=2[/tex]
From the normal distribution table, P(60.1 < x < 74.1) = P(-2 < z < 2) = P(z < 2) - P(z < -2) = 0.9772 - 0.0228 = 95.44%
95.44% of adults in the town have heights between 60.1 and 74.1 inches
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