IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual

IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. Find the probability that the person has an IQ score between 92 and 108.

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  1. (a) The distribution of X is N (100, 15²).

    (b) The probability that a person has an IQ greater than 130 is 0.0228.

    (c) The minimum IQ needed to qualify for the Mensa organization is 131.

    (d) The middle 20% of IQs fall between 96 and 104.

    Step-by-step explanation:

    The random variable X is defined as the IQ of an individual.

    (a)

    The random variable X is normally distributed with mean, μ = 100 and standard deviation, σ = 15.

    The probability density function of X is:

    [tex]f_{X}(x)=\frac{1}{15\sqrt{2\pi}}\times e^{-(x-100)^{2}/(2\times 15^{2})};\ -\infty[/tex]

    Thus, the distribution of X is N (100, 15²).

    (b)

    Compute the probability that a person has an IQ greater than 130 as follows:

    [tex]P(X130)=P(\frac{X-\mu}{\sigma}\frac{130-100}{15})[/tex]

                       [tex]=P(Z2)\\=1-P(Z[/tex]

    Thus, the probability that a person has an IQ greater than 130 is 0.0228.

    (c)

    Let x represents the top 2% of all IQs.

    Then, P (X > x) = 0.02.

    ⇒ P (X < x) = 1 - 0.02

    ⇒ P (Z < z) = 0.98

    The value of z is:

    z = 2.06.

    Compute the value of x as follows:

    [tex]z=\frac{x-\mu}{\sigma}\\2.06=\frac{x-100}{15}\\x=100+(2.06\times 15)\\x=130.9\\x\approx131[/tex]

    Thus, the minimum IQ needed to qualify for the Mensa organization is 131.

    (d)

    Let x₁ and x₂ be the values between which middle 20% of IQs fall.

    This implies that:

    [tex]P(x_{1}[/tex]

    The value of z is:

    z = 0.26.

    Compute the value of x as follows:

    [tex]-z=\frac{x_{1}-\mu}{\sigma}\\-0.26=\frac{x_{1}-100}{15}\\x_{1}=100-(0.26\times 15)\\x=96.1\\x\approx96[/tex]             [tex]z=\frac{x_{2}-\mu}{\sigma}\\0.26=\frac{x_{2}-100}{15}\\x_{2}=100+(0.26\times 15)\\x=103.9\\x\approx104[/tex]

    Thus, the middle 20% of IQs fall between 96 and 104.

  2. 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!!!

    [tex]IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual[/tex]
    [tex]IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual[/tex]
    [tex]IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual[/tex]

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