Review of Unit 6 Measurement:
1 answer:
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Answer:
a) The normal distribution function for the IQ of a randomly selected individual is presented in the first attached image to this solution.
b) The probability that the person has an IQ greater than 105 = P(X > 105) = 0.3707
The sketch of this probability is presented in the second attached image to this solution.
c) The minimum IQ needed to qualify for the Mensa organization = 130.81. Hence, P(X > 131) = 2%
The sketched image of this probability is presented also in the second attached image to the solution. The shaded region is the required probability.
d) The middle 40% of IQs fall between 92 and 108 IQ respectively.
P(x1 < X < x2) = P(92 < X < 108) = 0.40
The sketched image of this probability is presented in the third attached image to the solution. The shaded region is the required probability.
Step-by-step explanation:
The IQ of an individual is given as a normal distribution withh
Mean = μ = 100
Standard deviation = σ = 15
If X is a random variable which represents the IQ of an individual
a) The distribution of X will then be given as the same as that of a normal distribution.
f(x) = (1/σ√2π) {e^ - [(x - μ)²/2σ²]}
The normal distribution probability density function is more clearly presented in the first attached image to this question
b) Probability that the person has an IQ greater than 105.
To find this probability, we will use the normal probability tables
We first normalize/standardize 105.
The standardized score of any value is that value minus the mean divided by the standard deviation.
z = (x - μ)/σ = (105 - 100)/15 = 0.33
P(x > 105) = P(z > 0.33)
Checking the tables
P(x > 105) = P(z > 0.33) = 1 - P(z ≤ 0.33) = 1 - 0.6293 = 0.3707
The sketch of this probability is presented in the second attached image to this question. The shaded region is the required probability.
c) Mensa is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the Mensa organization.
We need to find x for P(X > x) = 2% = 0.02
Let the corresponding z-score for this probability be z'
P(X > x) = P(z > z') = 0.02
P(z > z') = P(z ≤ z') = 1 - 0.02 = 0.98
From the normal distribution table, z' = 2.054
z = (x - μ)/σ
2.054 = (x - 100)/15
x = 2.054×15 + 100 = 130.81 = 131 to the nearest whole number.
The sketched image of this probability is presented also in the second attached image to the solution. The shaded region is the required probability.
d) The middle 40% of IQs fall between what two values?
P(x1 < X < x2) = 0.40
Since the normal distribution is symmetric about the mean, the lower limit of this IQ range will be greater than the lower 30% region of the IQ distribution and the upper limit too is lesser than upper 30% region of the distribution.
P(X < x1) = 0.30
P(X > x2) = 0.30, P(X ≤ x2) = 1 - 0.30 = 0.70
Let the z-scores of x1 and x2 be z1 and z2 respectively.
P(X < x1) = P(z < z1) = 0.30
P(X ≤ x2) = P(z ≤ z2) = 0.70
From the normal distribution tables,
z1 = -0.524
z2 = 0.524
z1 = (x1 - μ)/σ
-0.524 = (x1 - 100)/15
x1 = -0.524×15 + 100 = 92.14 = 92 to the nearest whole number.
z2 = (x2 - μ)/σ
0.524 = (x2 - 100)/15
x2 = 0.524×15 + 100 = 107.86. = 108 to the nearest whole number.
The sketched image of this probability is presented in the third attached image to the solution. The shaded region is the required probability.
Hope this Helps!!!
