Question

Question 13 > B0/10 pts 596 Detail Use ONLY the Standard Normal Tables (Link) to answer the following... A set of exam scores is normally distributed and has a mean of 79.3 and a standard deviation of 7.2. What is the probability that a randomly selected score will be between 64 and 69? Answer = 0.5 (round to four decimal places) Note: Be careful...only use the Z Table here and round z scores to two places since that's what the table uses...do not use technology or the 68-95-99.7 Rule. Question Help: D Post to forum Submit Question

Answer 1

Solution :

Given, X follows Normal distribution with,

  $\mu$ = 79.3

  $\sigma$ = 7.2

Find, P(64< x< 69)

= P(X < 69) - P(X < 64)

=  P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596060608&ymreqid=69d777f6-ff7f-1f97-1c77-9d00da01eb00&sig=0edPHGcIw0wLAKfLIk48ng--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596060608&ymreqid=69d777f6-ff7f-1f97-1c77-9d00da01eb00&sig=mXNz0x84hCGNFQ9NK_g8Jw--~D$ <  (69 - 79.3)/7.2] - P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596060608&ymreqid=69d777f6-ff7f-1f97-1c77-9d00da01eb00&sig=0edPHGcIw0wLAKfLIk48ng--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596060608&ymreqid=69d777f6-ff7f-1f97-1c77-9d00da01eb00&sig=mXNz0x84hCGNFQ9NK_g8Jw--~D$ <  (64 - 79.3)/7.2]

= P[Z < -1.43] - P[Z < -2.12]

= 0.0764 - 0.017 ..Use z table

= 0.0594

P(64< x< 69) = 0.0594