Question

on any given day, 34% of sales at rubys jewelry store are from necklaces, while 28% of sales at nugget jewels are from necklaces. suppose rubys has 50 customers and nugget jewels has 60 customers on a randomly selected day. let $r =$ the proportion of sales that are from necklaces at rubys and $n =$ the proportion of sales that are from necklaces at nugget jewels. what is the probability that the proportion of sales from necklaces at rubys will be less than the proportion of sales from necklaces at nugget jewels?
find the z-table here.
○ 0.159
○ 0.250
○ 0.264
○ 0.750

Explanation:

Step1: Define given values

$p_R = 0.34$, $n_R = 50$, $p_N = 0.28$, $n_N = 60$
We need $P(R < N) = P(R - N < 0)$

Step2: Calculate mean of $R-N$

Mean $\mu_{R-N} = p_R - p_N = 0.34 - 0.28 = 0.06$

Step3: Calculate standard error

$$\begin{align*} SE_{R-N} &= \sqrt{\frac{p_R(1-p_R)}{n_R} + \frac{p_N(1-p_N)}{n_N}}\\ &= \sqrt{\frac{0.34(1-0.34)}{50} + \frac{0.28(1-0.28)}{60}}\\ &= \sqrt{\frac{0.34\times0.66}{50} + \frac{0.28\times0.72}{60}}\\ &= \sqrt{\frac{0.2244}{50} + \frac{0.2016}{60}}\\ &= \sqrt{0.004488 + 0.00336}\\ &= \sqrt{0.007848} \approx 0.0886 \end{align*}$$

Step4: Compute z-score

$z = \frac{0 - \mu_{R-N}}{SE_{R-N}} = \frac{0 - 0.06}{0.0886} \approx -0.677$

Step5: Find probability from z-table

Look up $z \approx -0.68$ in z-table, the corresponding left-tail probability is approximately 0.2483, closest to 0.250.

Answer:

0.250