Question

MAT305-02 UG Spring 2019 Session 1 Homework: Week 4 Homework Score: 0.4 of 1 pt Save | 1 of 10(1 complete) HW Score: 4%, 0 4 of 10. 6.2.2 Question Help Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete parts (a) through (d) below Click here to view page 1 of the cumulative standardized normal distribution table, a. What is the probability that Z is between-1.53 and 1.89? The probability that Z is between 1.53 and 1.89 is 9076 Round to four decimal places as needed.) b. What is the probability that Z is less than -1.53 or greater than 1.89 The probability that Z is less than 1.53 or greater than 1.89 is 0924 Round to four decimal places as needed.) C. What is the value of Z if only 5.5% of all possible Z values are larger? The value of Z if only 5.5% of all possible Z values are larger is 1 60 Round to two decimal places as needed) d. Between what two values of Z (symmetrically distributed around the mean) will 76.98% of all possible Z values be contained? The two values of Z for which 76.98% of all possible Z values are contained between are (Use ascending order. Round to two decimal places as needed.) and Enter your answer in the edit fields and then click Check Answer All parts showing 1:26 PM 1/30/2019 ^ -,

Answer 1

Part a) We are asked to find P( -1.53 < z< 1.89 )

We can use TI-84 calculator to find this probability.

Press 2NDkey - - > VARS key -- > select normalcdf() and hit enter .

Then plug the given values accordingly .

Scroll to paste and hit enter two times.

Answer : 0.9076

b) We are asked to find P( z <-1.53 or z >1.89  )

= 1 - P( -1.53 < z< 1.89 )

= 1- 0.9076

= 0.0924

c) We are asked to find z such that area to the right of that z is 5.5% or 0.055

Therefore area to the left of z is 1-0.055 = 0.945

We can use TI-84 calculator to find this z

Press 2NDkey - - > VARS key -- > select invNorm() and hit enter .Then plug the values accordingly.

For area : plug area to the left of z

Scroll to paste and hit enter two times.

Answer : 1.60

d) We are asked to find find two z scores such that area between then is 76.98% or 0.7698

The total area under the curve is 1, the middle area between z₁ and z₂ is 0.7689

Therefore area outside the z₁ and z₂ is 1 - 0.7689 = 0.2302

This area 0.2302 is equally distributed to both tail of the curve.

So area to the left of z₁ = 0.2302/2 = 0.1151 or area to the right of z₂ = 0.1151

We have to use function invNorm to find z₁ and z₂

For z₁ , area to the left is 0.1151

Scroll to paste and hit enter two times

Therefore z₁ = -1.20

Normal distribution is symmetric therefore z₂ = 1.20

The two values of z such that 76.98% area is between them is -1.20 and 1.20