Let z be a random variable with a standard normal
distribution.
Find “a” such that P(|Z| <A)= 0.95
This is what I have:
P(-A<Z<A) = 0.95
-A = -1.96
How do I use the symmetric property of normal distribution to make
A = 1.96?
My answer at the moment is P(|z|< (-1.96) = 0.95
Let z be a random variable with a standard normal distribution.
Find P(0 ≤ z ≤ 0.46), and shade the corresponding area under the
standard normal curve. (Use 4 decimal places.)
Z is a standard normal random variable, then k is ...
a. P(Z < k) = 0.92
b. P(Z > k) = 0.72
c. P(Z ≤ k) = 0.26
d. (A-Grade) P(−1 < Z < k) = 0.60
e. (A-Grade) P(k < Z < 1.7) = 0.57
f. (A-Grade) P(Z = k) = 0.00
Let z be a standard normal random variable with a mean of 0 and
a standard devi- ation of 1. Find the following probabilities:
(a) P(−0.5<z<0.5) (b) P(−.5<z<1.5)
(c) P(−1.5<z<−.75) (d) P(2<z<3)
Let the random variable Z follow a standard normal distribution,
and let Z1 be a possible value of Z that is representing the 90th
percentile of the standard normal distribution. Find the value of
Z1.
Find the following probabilities for the standard normal random
variable z z :
a) P(−2.07≤z≤1.93)= P ( − 2.07 ≤ z ≤ 1.93 ) =
(b) P(−0.46≤z≤1.73)= P ( − 0.46 ≤ z ≤ 1.73 ) =
(c) P(z≤1.44)= P ( z ≤ 1.44 ) =
(d) P(z>−1.57)= P ( z > − 1.57 ) =
1. If Z is a standard normal random variable, find
c such that P(−c ≤ Z ≤
c) = 0.82. [Answer to 2 decimal places]
2. Weakly earnings on a certain import venture are approximately
normally distributed with a known mean of $353 and unknown standard
deviation. If the proportion of earnings over $386 is 25%, find the
standard deviation. Answer only up to two digits after decimal.
3. X is a normal random variable with mean μ and
standard...
QUESTION 1
If Z is a standard normal random variable, then P(Z > 0)
=
0
1
0.4579
0.5
1 points
QUESTION 2
Company A claims that 20% of people in Sydney prefer its product
(Brand A). Company B disputes the 20% but has no idea whether a
higher or lower proportion is appropriate. Company B
randomly samples 400 people and 88 of them prefer Company A's
product (Brand A).
Assuming a 5% significance level, which one of the following...
Converting P(0.15 ≤ p ≤ 0.19) to the standard normal random
variable z for a sample of 500 households gives P(−1.19 ≤ z ≤
1.19). This is the probability that a sample of 500 households will
provide a sample proportion within 0.02 of the population
proportion, 0.17, of households that spend more than $100 per week
on groceries. Use a table to compute P(−1.19 ≤ z ≤ 1.19), rounding
the result to four decimal places. P(−1.19 ≤ z ≤ 1.19)...