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.

a. $X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

b. Find the probability that the person has an IQ greater than $120$. Include a sketch of the graph, and write a probability statement

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. Sketch the graph, and write the probability statement.

d. The middle $50\%$of IQs fall between what two values? Sketch the graph and write the probability statement.

Given in the question that

Mean$=100$

Standard deviation$=15$

$X=IQ$

Here we need to find$X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

Here mean is $100$

Standard deviation is $15$

So we can described the random variable $X$as

$X~N(100,15)$

Given in the question that

Mean$=100$

Standard deviation $=15$

$X=IQ$

Here we need to calculate the probability as,

$P(X>120)=1-P(X<120)$

$=1-P\left(\frac{X-\mu }{\sigma }<\frac{120-\mu }{\sigma }\right)$

$=1-P\left(Z<\frac{120-100}{15}\right)$

$=1-P(Z<1.3333)$

$=1-0.9088\left\{\begin{array}{l}\text{From standard}\\ \text{normal table}\end{array}\right\}$

$=0.0912$

So, It is found that the probability of a person that has an IQ greater than 120 is$0.0912$

From the above-calculated result the graph can be drawn below,

The probability statement will be$P(X>120)=0.0912$

Given in the question that

Mean$=100$

Standard deviation $=15$

$X=IQ$

Here the $2\%$of the preposition of people above the ${90}^{th}$percentile is same to each other

So, the ${90}^{th}$percentile be 'k'

the calculation will be,

$P(X>k)=1-P\left(\frac{X-\mu }{\sigma }<\frac{k-\mu }{\sigma }\right)$

$0.02=1-P\left(Z<\frac{k-100}{15}\right)$

$P\left(Z<\frac{k-100}{15}\right)=0.98$

$\frac{k-100}{15}=2.05\left\{\begin{array}{l}\text{From standard}\\ \text{normal table}\end{array}\right\}$

$k-100=2.05\times 15$

$k=30.75+100$

$k=130.75$

From the result, it is clear that the minimum IQ needed to qualify for the MENSA organization is $130.75$

The graphical presentation for the above result is given below,

Therefore the probability statement will be$P(X>k)=P(X>130.75)=0.02$

Given in the question that

Mean$=100$

Standard deviation$=15$

$X=IQ$

Here the middle $50\%$of value lie between ${25}^{th}$percentile and ${75}^{th}$percentile.

So the value can be calculated using the Ti-83 calculator

For that, use invNorm( ) function and enter the given information to find the percentiles. The screenshot is given as below:

Therefore the middle $50\%$of IQs fall between$89.88$and$110.12$

The graph for the above result is given below,

Therefore the probability statement is$P(89.88<X<110.12)=0.5001$.