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4 years ago

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In a study of environmental lead exposure and IQ, the data was collected from 148 children in Boston, Massachusetts. Their IQ sc

ores at age of 10 approximately follow a normal distribution with mean of 115.9 and standard deviation of 14.2. Suppose one child had an IQ of 74. The researchers would like to know whether an IQ of 74 is an outlier or not.
Calculate the lower fence for the IQ data, which is the lower limit value that the IQ score can be without being considered an outlier. Keep a precision level of two decimal places for the lower fence.

1 answer:

serious [3.7K]4 years ago

8 0

Answer:

a) Lower inner fence = 77.6168 = 77.62 to 2 d.p

Lower outer fence = 48.9044 = 48.90 to 2 d.p

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159

Step-by-step explanation:

Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Q₁ = 25th percentile = lower quartile

IQR = Inter quartile Range = Q₃ - Q₁

Q₃ = 75th percentile = upper quartile

To calculate Q₁ for a normal distribution with only mean and standard deviation known,

We need the standardized score whose probability is 0.25 P(z) = 0.25

From the normal distribution table

z = (± 0.674)

z = (x - xbar)/σ

x = the value in the data we're interested in,

xbar = mean = 115.9

σ = standard deviation = 14.2

Lower quartile corresponds to (z = - 0.674)

- 0.674 = (x - 115.9)/14.2

Q₁ = X = 106.3292

The upper quartile, Q₃ corresponds to z = (+0.674)

Q₃ = 125.4708

IQR = 125.4708 - 106.3292 = 19.1416

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168

Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)

We standardize 74 by obtaining its z-score

z = (x - xbar)/σ

z = (74 - 115.9)/14.2 = - 2.95

P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)

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