AP Statistics 5.2 The Normal Distribution, Revisited - MCQs - Exam Style Questions

Question

The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean $80$ and standard deviation $7$. Volunteers in the top $20$ percent of hours worked will receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than $90$?

(A) $0.077$
(B) $0.123$
(C) $0.134$
(D) $0.618$
(E) $0.923$

▶️ Answer/Explanation

Detailed solution

Let $X\sim \mathcal N(80,7)$ be hours worked. “Top $20\%$” cutoff is the $80^\text{th}$ percentile $c$ where $P(X>c)=0.20\Rightarrow P(X<c)=0.80$.
Using technology/tables, $c\approx 85.89$ hours (since $z_{0.80}\approx 0.8416$ and $80+0.8416\times 7\approx 85.89$).
We want $P(\text{merit}\mid X<90)=P(X>85.89\mid X<90)=\dfrac{P(85.89<X<90)}{P(X<90)}$.
Convert bounds to $z$-scores: $z_1=\dfrac{85.89-80}{7}\approx 0.841$, $z_2=\dfrac{90-80}{7}\approx 1.429$.
From normal CDF values: $\Phi(z_1)\approx 0.800$, $\Phi(z_2)\approx 0.923$.
Then $P(85.89<X<90)=\Phi(z_2)-\Phi(z_1)\approx 0.923-0.800=0.123$.
And $P(X<90)=\Phi(z_2)\approx 0.923$.
Therefore \[ P(\text{merit}\mid X<90)=\frac{0.123}{0.923}\approx 0.134. \] ✅ Answer: (C)

Question

Scientists working for a water district measure the water level in a lake each day. The daily water level in the lake varies due to weather conditions and other factors. The daily water level has a distribution that is approximately normal with mean water level of 84.07 feet. The probability that the daily water level in the lake is at least 100 feet is 0.064.

Which of the following is closest to the probability that on a randomly selected day the water level in the lake will be at least 90 feet?

(A) 0.29
(B) 0.31
(C) 0.34
(D) 0.37
(E) 0.50

▶️ Answer/Explanation

Detailed solution

1. Find the Standard Deviation ($\sigma$):
We are given $\mu = 84.07$ and $P(X \ge 100) = 0.064$. The area to the left of 100 is $P(X < 100) = 1 – 0.064 = 0.936$.
We find the z-score corresponding to this cumulative area, which is $z \approx 1.52$.
Now, we use the z-score formula $z = \frac{x – \mu}{\sigma}$ to solve for $\sigma$:
$1.52 = \frac{100 – 84.07}{\sigma} \Rightarrow \sigma = \frac{15.93}{1.52} \approx 10.48$

2. Calculate the Z-score for 90 feet:
Using the calculated $\sigma$, we find the z-score for $x = 90$.
$z = \frac{90 – 84.07}{10.48} = \frac{5.93}{10.48} \approx 0.5658$
We can round this to $z \approx 0.57$.

3. Find the Desired Probability:
We need to find $P(X \ge 90)$, which corresponds to $P(Z \ge 0.57)$.
Using a standard normal table, $P(Z < 0.57) \approx 0.7157$.
Therefore, $P(Z \ge 0.57) = 1 – P(Z < 0.57) \approx 1 – 0.7157 = 0.2843$.
This value is closest to 0.29.
✅ Answer: (A)

AP Statistics 5.2 The Normal Distribution, Revisited - MCQs - Exam Style Questions

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