Question

a large corporation sponsors a bowling league for its women employees. their mean score was 144 pins with a standard deviation of 12 pins. note: bowling scores are normally distributed. 1) what is the probability that a women’s score is greater than 167 pins? 2) if x and y represent two different women in the league, what is the probability that the sum of their scores is greater than 300 pins? 3) if x and y represent two different women in the league, what is the probability that the difference in their scores is less than 10 pins?

Explanation:

1) Probability a score > 167 pins

Step1: Calculate z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 167$, $\mu=144$, and $\sigma = 12$.
$z=\frac{167 - 144}{12}=\frac{23}{12}\approx1.92$

Step2: Find $P(Z>1.92)$

We know that $P(Z > z)=1 - P(Z\leq z)$. From the standard normal table, $P(Z\leq1.92) = 0.9726$. So $P(Z > 1.92)=1 - 0.9726 = 0.0274\approx0.0275$ (or 0.0276 as in the given sketch)

Step1: Find the distribution of $X + Y$

If $X\sim N(\mu_{X},\sigma_{X}^{2})$ and $Y\sim N(\mu_{Y},\sigma_{Y}^{2})$ and $X$ and $Y$ are independent, then $X + Y\sim N(\mu_{X}+\mu_{Y},\sigma_{X}^{2}+\sigma_{Y}^{2})$. Here, $\mu_{X}=\mu_{Y}=144$, $\sigma_{X}=\sigma_{Y}=12$. So $\mu_{X + Y}=144 + 144=288$ and $\sigma_{X + Y}^{2}=12^{2}+12^{2}=288$, so $\sigma_{X + Y}=\sqrt{288}\approx16.97$

Step2: Calculate z - score for $x = 300$

Using the z - score formula $z=\frac{(X + Y)-\mu_{X + Y}}{\sigma_{X + Y}}$, we have $z=\frac{300 - 288}{16.97}=\frac{12}{16.97}\approx0.71$

Step3: Find $P(X + Y>300)$

$P(X + Y>300)=1 - P(X + Y\leq300)=1 - P(Z\leq0.71)$. From the standard normal table, $P(Z\leq0.71)=0.7611$. So $P(X + Y>300)=1 - 0.7611 = 0.2389$

Step1: Find the distribution of $X - Y$

If $X\sim N(\mu_{X},\sigma_{X}^{2})$ and $Y\sim N(\mu_{Y},\sigma_{Y}^{2})$ and $X$ and $Y$ are independent, then $X - Y\sim N(\mu_{X}-\mu_{Y},\sigma_{X}^{2}+\sigma_{Y}^{2})$. Here, $\mu_{X}-\mu_{Y}=144 - 144 = 0$ and $\sigma_{X - Y}^{2}=12^{2}+12^{2}=288$, so $\sigma_{X - Y}=\sqrt{288}\approx16.97$

Step2: Calculate z - scores for $x=- 10$ and $x = 10$

We want $P(- 10For $x=-10$: $z_{1}=\frac{-10-0}{16.97}\approx - 0.59$
For $x = 10$: $z_{2}=\frac{10 - 0}{16.97}\approx0.59$

Step3: Find $P(-0.59

$P(-0.59

Answer:

The probability is approximately $0.0276$ (or $0.0274$)

2) Probability the sum of two scores > 300 pins