9The middle area between z1 and z2 is 0.7500 P(z1<z<z2) - 0.7500 1.04 1.08 1.15 1.04...
QUESTION 18 Find the area P(-2.21 < z < 0) under the standard normal curve. 1.1050 0.4864 0.0136 O-0.9864
Data analysis 5. (10 points) Please determine the following probability given the Z value using the standard normal distribution table a) P(Z < 1.28) b) P(Z>1.45)
(10 pts.) 6a) Find P.21 <z < 1.06) 6b) Find a z-score satisfying the condition that 75% of the total area is to the left of z. 6c) Find a z-score satisfying the condition that 80% of the total area is to the right of z. REFER to the table.
Question completion status: QUESTION 9 The shaded area represents the probability: 1.68 OP (z < 1.68) P(-1.37>2 > 1.68) P(-1.37<z> 1.68) P(-1.37 <2<1.68)
Find: P(-2.36 < Z < -1.04 ) using the standard normal distribution table. O a. 1401 b..0717 C..1583 d. 9066 e. 8417 Of..0934
4.28 If Z ~ N(0,1), find the following probabilities: a. P(Z <1.38) b. P(Z > 2.14) c. P(-1.27 <Z<-0.48)
ULULUI Let Z be a standard normal random variable. What is P(-2.22 <Z<0.25)? 0.2212 0.3488 0.5855 O 0.6902
15. Let X1, . . . , Xn be id from pmf p(z; θ)-(1-0)"-10; ;z=1,2, 3, ,and 0 < θ < 1. (a) Find the maximum likelihood estimator of θ (b) Find the maximum likelihood estimate of θ using the observed sample of 5,8,11.
IS Find the following probability for the standard normal random variable z. a. P(z<-1.02) b. P(z <2.03) c. P(0.68 szs2.03) d. P(-2.66szs1.56) a. P(z -1.02)(Round to four decimal places as needed.) b. Pize 2.03)=[] (Round to four decimal places as needed.) ook c. P(0.68 szs2.03) (Round to four decimal places as needed.) d.P(-2.66 s zs 1.56) = [□ (Round to four decimal places as needed.)
Problem 7.6 The annular surface 1 (cm) < p < 3 cm), z = 0, carries the non-uniform surface charge density ps = 5p (nC/m²). Find V at P(0,0,2 cm) if V = 0 at infinity.