1.) Find the value of k for which the given function is a probability density fu

Question

1.) Find the value of k for which the given function is a probability density function.

f(x) = kekx on [0, 8]

k=

2.) This exercise uses the normal probability density function and requires the use of either technology or a table of values of the standard normal distribution.

Repeated measurements of a student's IQ yield a mean of 120, with a standard deviation of 5. What is the probability that the student has an IQ between 117 and 123? (Round your answer to four decimal places.)

3.) This exercise uses the normal probability density function and requires the use of either technology or a table of values of the standard normal distribution.

The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.89 per access line per month, with a standard deviation of $2.75. Company A's operating expenses were $29.00 per access line per month. Assuming a normal distribution of operating expenses, estimate the percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean. (Round your answer to two decimal places.)
________ %

4.)This exercise uses the normal probability density function and requires the use of either technology or a table of values of the standard normal distribution.

The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.95 per access line per month, with a standard deviation of $2.45. Company N's operating expenses were $36.52 per access line per month in the first half of 1994. Estimate the percentage of regional phone companies whose operating expenses were higher than those of Company N. (Round your answer to two decimal places.)

________%

Explanation / Answer

Solution

Back-up Theory

If a continuous random variable, X, has pdf (probability density function) f(x), for a ? x ? b, and 0 otherwise, then

f(x) must satisfy 3 conditions:

1. f(x) ? 0      (2). f(x) ? 1      (3) integral from ‘a’ to ‘b’ of {f(x)} = 1. ……………………(A)

Establishing conditions (1) and (3) necessarily implies condition (2) ……………………(B)

If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then, pdf of X, f(x) = {1/??(2?)}e^-[(1/2){(x - µ)/?}2] …………………………….(A)

Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X ? or ? t) = P[{(X - µ)/? } ? or ? {(t - µ)/? }] = P[Z ? or ? {(t - µ)/?}] .………(2)

X bar ~ N(µ, ?2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar ? or ? t) = P[Z ? or ? {(?n)(t - µ)/? }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be found using Excel Function……………………..(5)

Q1

If the function f(x) = kekx on [0, 8] is a probability density function, then vide (A) and (B) above,

Integral kekx on [0, 8] must be 1 assuming k is non-negative,

i.e., (ekx) [0, 8] = 1 or

e8k – 1 = 1 or

e8k = 2 or

8k = ln2 or

k = (ln2)/8

   = 0.693147/8

   = 0.086643 ANSWER

Q2

Let X = Student's IQ. We assume X ~ N(120, 52).

Probability that the student has an IQ between 117 and 123 = P(117 < X < 123)

= P[{(117 - 120)/5} ? Z ? {(123 - 120)/5}] [vide (2) above]

= P(- 0.6 ? Z ? 0.6)

= P(Z ? 0.6) - P(Z ? 0.6)

= 0.7257 – 0,2743 [from Standard Normal Probability Table]

= 0.4514 ANSWER

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