Question

1.

Assume that the average weight of an NFL player is 245.7 lbs with a standard deviation of 34.5 lbs. The distribution of NFL weights is not normal. Suppose you took a random sample of 32 players.

What is the probability that the sample average will be between 242 and 251 lbs? Round your answer to three decimal places, eg 0.192.

2.

Given a sample with a mean of 57 and a standard deviation of 8, calculate the following probabilities using Excel. Notice this is not a standard normal distribution. Round your answer to three decimals points, e.g. 0.753.

Pr(X < 59.5)

3. If Z~N(0,1), what is Pr(Z > -2.44)? Round your answer to three decimals, e.g. 0.491.

4. A continuous random variable X has a normal distribution with mean 22. The probability that X takes a value less than 25.5 is 0.64. Use this information and the symmetry of the density function to find the probability that X takes a value greater than 18.5. Enter your answer as a number rounded to two decimal points, e.g. 0.29.

5. Z is distributed as a standard normal variable. Find the value of a such that Pr (Z < a) = 0.409. Round your answer to three decimal places, e.g. 2.104.

6.

Given a sample with a mean of 196 and a standard deviation of 20, calculate the following probability using Excel. Note the sign change. Round your answer to three decimal places, e.g. 0.491.

Pr(172 < X < 225)

7. If Z~N(0,1), calculate Pr(0.27 < Z < 1.08). You can use Excel. Round your answer to three decimal places, e.g. 0.491

Answer 1

1.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 245.7
standard Deviation ( sd )= 34.5/ Sqrt ( 32 ) =6.0988
sample size (n) = 32
probability that the sample average will be between 242 and 251 lbs
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 242) = (242-245.7)/34.5/ Sqrt ( 32 )
= -3.7/6.0988
= -0.6067
= P ( Z <-0.6067) From Standard Normal Table
= 0.272
P(X < 251) = (251-245.7)/34.5/ Sqrt ( 32 )
= 5.3/6.0988 = 0.869
= P ( Z <0.869) From Standard Normal Table
= 0.8076
P(242 < X < 251) = 0.8076-0.272 = 0.536
2.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 57
standard Deviation ( sd )= 8
P(X < 59.5) = (59.5-57)/8
= 2.5/8= 0.313
= P ( Z <0.313) From Standard Normal Table
= 0.623
3.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0
standard Deviation ( sd )= 1
P(X > -2.44) = (-2.44-0)/1
= -2.44/1 = -2.44
= P ( Z >-2.44) From Standard Normal Table
= 0.993
4.
A continuous random variable X has a normal distribution with mean 22.
The probability that X takes a value less than 25.5 is 0.64.
z value from p value , from Z table
standard deviation = (25.5-22)/0.36 =9.722
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 22
standard Deviation ( sd )= 9.722
P(X > 18.5) = (18.5-22)/9.722
= -3.5/9.722 = -0.36
= P ( Z >-0.36) From Standard Normal Table
= 0.64
5.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0
standard Deviation ( sd )= 1
P ( Z < x ) = 0.409
Value of z to the cumulative probability of 0.409 from normal table is -0.2301
P( x-u/s.d < x - 0/1 ) = 0.409
That is, ( x - 0/1 ) = -0.2301
--> x = -0.2301 * 1 + 0 = -0.2301
6.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 196
standard Deviation ( sd )= 20
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 172) = (172-196)/20
= -24/20 = -1.2
= P ( Z <-1.2) From Standard Normal Table
= 0.115
P(X < 225) = (225-196)/20
= 29/20 = 1.45
= P ( Z <1.45) From Standard Normal Table
= 0.926
P(172 < X < 225) = 0.926-0.115 = 0.811
7.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0
standard Deviation ( sd )= 1
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.27) = (0.27-0)/1
= 0.27/1 = 0.27
= P ( Z <0.27) From Standard Normal Table
= 0.606
P(X < 1.08) = (1.08-0)/1
= 1.08/1 = 1.08
= P ( Z <1.08) From Standard Normal Table
= 0.86
P(0.27 < X < 1.08) = 0.86-0.606 = 0.254