Download 1. x has a normal distribution with a mean of 80.0 and a standard

1. x has a normal distribution with a mean of 80.0 and a standard deviation of 4.0. find
the following probabilities:
a. p(x<78.0)
z = (78 – 80) / 4 = -0.5
From an online statistics calculator:
p(z < -0.5) = 0.3085
Therefore, p(x < 78.0) = p(z < -0.5) = 0.3085
b. p(78.0<x<87.0)
As above, the z-score for x = 78 is -0.5
The z-score for x = 87.0 is (87 – 80) / 4 = 1.75
Again, from the calculator:
p(z < -0.5) = 0.3085
p(z < 1.75) = 0.9599
The probability of x being between the two limits is equal to the
difference:
p(78.0 < x < 87.0) = p(z < 1.75) – p(z < -0.5)
= 0.9599 – 0.3085 = 0.6514
2. Answer the following:
a. find the binomial probability p(x=6), where n=15 and p =0.60.
p = 0.6,
q = 1 – p = 1 – 0.6 = 0.4
p(x = 6) = C(15, 6)(0.6)6(0.4)15-6 = 5005(0.6)6(0.4)9
= 5005(0.046656)(0.0002621)
= 0.0612
b. set up without solving the binomial probability p(x is at most 6) using probability
notation.
The probability that x is at most six is equal to the sum of the probabilities that
x is exactly any integer value less than or equal to 6.
p(x ≤ 6) = p(x=0) + p(x=1) + p(x=2) + p(x=3) + p(x=4) + p(x=5) + p(x=6)
c. How would you find the normal approximation to the binomial probability p(x=6)in
part a? please show how you would calculate the mean and the standard deviation in the
formula for the normal approximation to the binomial and show the final formula you
would use without going through all the calculations.
The mean is equal to np = (15)(0.6) = 9.0
The standard deviation is equal to √(npq) = √(15*0.6*0.4) = √3.6 = 1.897
The probability of exactly 6 successes is approximated by the area under the
normal curve between 6.5 and 5.5:
p(x = 6) = p(x < 6.5) – p(x < 5.5)
Calculate the z-score for both limits:
x < 6.5:
(6.5 – 9.0) / 1.897 = -1.3179
x < 5.5:
(5.5 – 9.0) / 1.897 = -1.8450
The normal approximation is then p(z < -1.3179) – p(z < -1.8450)
3. If the random variable z is the standard normal score, is it true that p(-1<z<1) could
easily be approximated without referring to a table? why or why not.
Yes, this is true. A z-score of 1 indicates a point one standard deviation greater
than the mean, while a z-score of -1 indicates a point one standard deviation less
than the mean. The Empirical Rule tells us that the area under the normal curve
within one standard deviation of the mean is 68%. Then p(-1 < z < 1) is 0.68.
4. find the area under the standard normal curve for the following:
a. p(z<-0.74)
Using an online calculator: p(z < -0.74) = 0.2297
b. p(-0.87<z<0)
p(-0.87 < z < 0) = p(z < 0) – p(z < -0.87) = 0.5 - 0.1922 = 0.3078
c. p(-2.03<z<1.66)
p(-2.03<z<1.66) = p(z < 1.66) – p(z < -2.03) = 0.9515 - 0.0212 = 0.9303
4. find the value of z such that approximately 46.71% of the distribution lies between it
and the mean.
The area under the normal curve corresponding to 46.71% of the distribution is
equal to 0.4671.
Half of this area is to the left of the mean, and half of the area is to the right.
The total area to the left of the upper limit will be:
0.5 + (0.4671 / 2) = 0.73355
(The upper limit is calculated this way because it fits with the use of the statistical
calculator. If using tables, then find the z-value that corresponds to an area of
0.4671 ÷ 2 = 0.23355)
z = 0.6236
This means that 46.71% of the distribution will lie between z = -0.6236 and z =
0.6236.
On the other hand, if the intent of this problem was merely to have you use the
table to find the z-score for an area to the right of the mean equal to 0.4671, then
find 0.4671 in the table and read the corresponding z-value. (It may be necessary
to interpolate.) The answer in this case is z = 1.84.