Question

A sample of final exam scores is normally distributed with a mean equal to 23 and a variance equal to 16.

• Part (a)

  What percentage of scores are between 19 and 27? (Round your answer to two decimal places.)

• Part (b)

  What raw score is the cutoff for the top 10% of scores? (Round your answer to one decimal place.)

• Part (c)

  What is the proportion below 18? (Round your answer to four decimal places.)

• Part (d)

  What is the probability of a score less than 28? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 23

variance = 16

standard deviation = = 4

(a)

P(19 < x < 27) = P[(19 - 23)/ 4) < (x - ) /  < (27 - 23) / 4) ]

= P(-1 < z < 1)

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

= 0.8413 - 0.1587

= 0.6826 = 68.26%

percentage = 68.26%

(b)

Using standard normal table ,

P(Z > z) = 10%

1 - P(Z < z) = 0.1

P(Z < z) = 1 - 0.1

P(Z < 1.28) = 0.9

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 4 + 23 = 28.1

scores = 28.1

(c)

P(x < 18) = P[(x - ) / < (18 - 23) / 4]

= P(z < -1.25)

= 0.1056

Proportion = 0.1056

(d)

P(x < 28) = P[(x - ) / < (28 - 23) / 4]

= P(z < 1.25)

= 0.8944

probability = 0.8944