Question

Question:

a) Scores for a particular standardized test are normally distributed with a mean of 80 and a standard deviation of 14.

Find the probability that a randomly chosen score is;

i. No greater than 70

ii. At least 95

iii, Between 70 and 95

iv.. A student was told that her percentile score on this exam is 72% . Approximately what is her raw score?

1. b) If Z∼N(0,1) , find the following probabilities:   i, Ρ(−2.56 <Ζ<−0.134) ii. Ρ(−1.762 <Ζ<−0.246)   iii. Ρ(0 <Ζ<−1.73) iv. P( lΖl <−1.78)     v. ( lΖl < 2.326)

Answer 1

Solution-A:

P(X<=70)

z=x-mean/sd

=70-80/14

=-10/14

z =-0.7142857

P(Z<-0.7142857)

from standard normal tables

=0.2375

P(X<=70)=0.2375

Solution-b:

P(X>=95)

z=95-80/14

z=1.071429

P(Z>1.071429)

=1-P(Z<1.071)

=1-0.858

=0.142

P(At least 95)=0.1420

Solutionc:

P(70<X<90)

P(-0.7142857<Z<1.071429)

P(Z<1.071429)-P(Z<-0.7142857)

=0.858-0.2375

=0.6205

P(Between 70 and 95)=0.6205

SolutionD:

z score for 72 percentile is

0.5828415 which we got using in R :qnorm(0.72)
z=x-mean/sd

0.5828415=x-80/14

x=0.5828415*14+80

x=88.15978

her raw score is 88