Let Z be the standard normal random variable. What is P(Z<2.38)? 0.5239 0.7764 0.9162 0.9913 None...

Let Z be the standard normal random variable. What is P(Z<2.38)?

		0.5239
		0.7764
		0.9162
		0.9913
		None of the above

Solutions

Expert Solution

orchestra answered 1 year ago

Next > < Previous

Related Solutions

Let z be a random variable with a standard normal distribution. Find “a” such that P(|Z|...

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 ≤...

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) =...

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...

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...

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...

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...

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...

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...

If P(-2 < Z < k)=.6 , where Z is a standard normal random variable, then...

If P(-2 < Z < k)=.6 , where Z is a standard normal random variable, then k is... Select one: a. 0.195 b. 0.73 c. 0.55 d. -0.40

Converting P(0.15 ≤ p ≤ 0.19) to the standard normal random variable z for a sample...

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)...

Subjects