Respuesta :
Answer:
The height that cuts off the top 5% is 74.83 inches.
Step-by-step explanation:
We are given that in the survey, respondents were grouped by age. In the 20-29 age group, the heights were normally distributed, with a mean of 69.9 inches and a standard deviation of 3.0 inches.
Let X = heights of respondents
So, X ~ N([tex]\mu=69.9,\sigma^{2} =3^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{ X -\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean height = 69.9 inches
[tex]\sigma[/tex] = standard deviation = 3.0 inches
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, we have to find the height that cuts off the top 5%, that means;
P(X > [tex]x[/tex]) = 0.05 {where [tex]x[/tex] is the height that cuts off top 5%}
P( [tex]\frac{ X -\mu}{\sigma}[/tex] > [tex]\frac{ x -69.9}{3}[/tex] ) = 0.05
P(Z > [tex]\frac{ x -69.9}{3}[/tex] ) = 0.05
Now, in the z table the critical value of X that gives the area of top 5% is given as 1.6449.
So, [tex]\frac{ x -69.9}{3} = 1.6449[/tex]
[tex]x -69.9= 1.6449 \times 3[/tex]
[tex]x[/tex] = 69.9 + 4.9347 = 74.83
Hence, the height that cuts off the top 5% is 74.83 inches.
