Question

1.Consider the probability that no more than 1919 out of 157157 people have been in a car accident. Assume the probability that a given person has been in a car accident is 15%15%.

Approximate the probability using the normal distribution. Round your answer to four decimal places.

2. Scores on a test have a mean of 78.878.8 and 77 percent of the scores are above 8989. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth, if necessary.

Answer 1

Q1: Proportion, p = 0.15

Sample size, n = 157

Mean, µ = n*p = 157 * 0.15 = 23.55

Standard deviation, σ = √(n*p*(1-p)) = √(157 * 0.15 * 0.85) = 4.4741

P(X ≤ 19) =

Using continuity correction :

= P(X ≤ 19+0.5)

= P((X - µ)/σ ≤ (19.5 - 23.55)/4.4741)

= P(z ≤ -0.9052)

Using excel function:

= NORM.S.DIST(-0.9052, 1)

= 0.1827

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Q2. P(x > a) = 0.07

= 1 - P(x < a) = 0.07

= P(x < a) = 0.93

Z score at p = 0.93 using excel = NORM.S.INV(0.93) = 1.4758

σ = (X - µ)/z = (89 -78.8)/1.4758 = 6.9