2.) Suppose we have population data for the age at which children say their firs

Question

2.) Suppose we have population data for the age at which children say their first word. The data is normally distributed, and the average age is 286 days, with a standard deviation of 27. Given this information, answer the following questions. Additionally, for each question: Draw a sketch of the normal distribution curve for each problem, including the z-score and raw-score scales, and shade in the regions described in the question. Name the column of the z-table that you use (if applicable). A) Suppose we have a baby who said their first word at 221 days. What is their percentile rank (i.e., what percent of the population spoke at this age or younger)? (16 pts) B) We are interested in the age cut-off for the 80th percentile (i.e., the raw score for age that corresponds to the youngest 80% of the population) for age at which children say their first word. Find this age. (16 pts) c) we are interested in the scores that serve as boundaries for the middle 50% of the distribution (i.e., between the 25h percentile (Q1end 75h percentile (.e., Q3). Find these ages. (28 pts) 3.) Mathematics SAT scores are normally distributed with a mean of 500 and a standard deviation of 110. Given this information, answer the following questions. Additionally, for each question: ry .Draw a sketch of the normal distribution curve for each problem, including the z score and raw-score ry scales, and shade in the regions described in the question. .Name the column of the z table that you use (if applicable). A) what Math SAT score separates the bottom 90% from the top 1062 Shade in the bottom portion ie of the distribution.) (16 pts) B) What Z-scores separate the middle 95% from the extreme 587 (These z-scores separate the middle com o n 95% from the more extreme, uncom mo n" 5:2.5% in the uppertal, and 2.51 in the lower tail. These are known as the "critical z" scores, or " stit", and you will see these again in a few weeks when we visit hypothesis testing and z tests.) (Shade in the tails of the Aa ! F8 F6

Explanation / Answer

(2)

Data given:

Mean, m = 286

Standard Deviation, S = 27

(a)

At X = 221, the corresponding z-score is:

z = (X-m)/S = (221-286)/27 = -2.407

The corresponding p-value for this z-score is:

p = 0.00804

So, percentile rank = 0.00804*100 = 0.804%

(b)

At p = 0.80, the corresponding z-score is:

z = 0.841 = (X-m)/S

So,

(X-286)/27 = 0.841

Solving we get:

X = 308.7

(c)

At p = 0.25, the corresponding z-score is:

z = -0.674 = (X-m)/S

So,

(X-286)/27 = -0.674

Solving we get:

X = 267.8

At p = 0.75, the corresponding z-score is:

z = 0.674 = (X-m)/S

So,

(X-286)/27 = 0.674

Solving we get:

X = 304.2

So the required range is 267.8 to 304.2

Hope this helps !

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