Question

Exam grades: Scores on a statistics final in a large class were normally distributed with a mean of 79 and a standard deviation of 8. Use the TI-84 PLUS calculator to answer the following. Round the answers to at least four decimal places. (a) What proportion of the scores were above 93? (b) What proportion of the scores were below 66? (c) What is the probability that a randomly chosen score is between 70 and 90?

Answer 1

Given that ,

mean = $\mu$ = 79

standard deviation = $\sigma$ = 8

a) P(x >93 ) = 1 - p( x< 93 )

=1- p P[(x - $\mu$ ) / $\sigma$ < (93-79) /8 ]

=1- P(z < 1.75 )

= 1 - 0.9599 = 0.0401

probability = 0.0401

b)

P(x < 66 ) = P[(x - $\mu$ ) / $\sigma$ < (66-79) /8 ]

= P(z < -1.625 )

= 0.0521

probability = 0.0521

c)

P( 70< x < 90 ) = P[(70-79) /8 ) < (x - $\mu$ ) /$\sigma$  <(90-79) /8 ]

= P( -1.125< z < 1.375)

= P(z <1.375 ) - P(z < -1.125)

= 0.9154 - 0.1303 = 0.7851

Probability = 0.7851