Question

X is a random variable which is normally distributed with a mean of 99.01 and a standard deviation of 15.56. Use the Excel function NORMINV to determine the required value of Xo to two decimal places. Give your answer in the form xx.xx. P(X < Xo) = 0.0344

Answer:

Answer 1

X is random variable whuwhis normally distributed with mean 99.01 and standard deviation 15.56. Now we use excel function NORMINV

to determine the required value of X_{​​​​​​0} such that P(X<X_{​​​​​​0}​​​)=0.0344.

Syntax for NORMINV function:-

NORMINV(probability, mean, standard_dev)

Here,

probability=0.0344 , mean=99.01 , standard_dev=15.56

i.e syntax is,

NORMINV(0.0344, 99.01, 15.56)

And using this function we get the value of X_{​​​​​​0}​​​​​= 70.69

REVIEW INSERT PAGE LAYOUT FORMULAS VIEW HOME DATA FILE A A Wrap Text 11 E Copy Paste A I U Merge & Center Format Painter Clipboard Font Alignment fr SUM X =NORMINV ( 0.0344, 99.01, 15.56) A C E F G H 1NORMINV(0.0344, 99.01, 15.56) 2 C3 7 8 9 10 11 12 13 14

PAGE LAYOUT DATA FORMULAS REVIEW HOME INSERT FILE XCut A A Calibri 11 自Copy Paste B I Format Painter Clipboard Alignment Font -NORMINV(0.0344, 99.01, 15.56) A1 E F H 70.69499 2 7 8 10 11 12 12 X CO