Suppose the scores on a chemistry test were normally distributed with a mean of 78 and a standard deviation of 10. If a student who completed the test is chosen at random, a.Find the probability that the student earned between 80 and 90 points. b.Find the probability that the student earned either less than 80 points or more than 90 points.

a) 0.30567

b) 0.69433

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

,a.Find the probability that the student earned between 80 and 90 points.

For 80 students

z = 80 - 78/10

= 0.2

P-value from Z-Table:

P(x = 80) = 0.57926

For 90 students

z = 90 - 78/10

=1.2

P-value from Z-Table:

P(x = 90) = 0.88493

The probability that the student earned between 80 and 90 points.

P(x = 90) - P(x = 80)

= 0.88493 - 0.57926

= 0.30567

b.Find the probability that the student earned either less than 80 points or more than 90 points.

For less than80 students

z = 80 - 78/10

= 0.2

P-value from Z-Table:

P(x < 80) = 0.57926

For 90 students

z = 90 - 78/10

=1.2

P-value from Z-Table:

P(x<90) = 0.88493

P(x>90) = 1 - P(x<90) = 0.11507

The probability that the student earned either less than 80 points or more than 90 points.

= 0.57926 + 0.11507

= 0.69433

ghjmutmum

step-by-step explanation:

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