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let random variable ( y ) represent the number of interviews conducted …

Question

let random variable ( y ) represent the number of interviews conducted for job openings at a certain company. the following table shows the cumulative probability distribution of the discrete random variable ( y ).

( y )( p(y leq y) )
60.2
70.4
80.6
90.8
101.0

khaleed claims that the distribution of ( y ) is skewed to the left with mean equal to 8 interviews. is khaleed’s claim correct?

Explanation:

Step1: Find individual probabilities

For discrete random variables, $P(Y=y) = P(Y \leq y) - P(Y \leq y-1)$:

  • $P(Y=5)=0-0=0$
  • $P(Y=6)=0.2-0=0.2$
  • $P(Y=7)=0.4-0.2=0.2$
  • $P(Y=8)=0.6-0.4=0.2$
  • $P(Y=9)=0.8-0.6=0.2$
  • $P(Y=10)=1.0-0.8=0.2$

Step2: Calculate the mean $\mu$

Mean formula: $\mu = \sum y \cdot P(Y=y)$

$$\begin{align*} \mu&=(5 \cdot 0)+(6 \cdot 0.2)+(7 \cdot 0.2)+(8 \cdot 0.2)+(9 \cdot 0.2)+(10 \cdot 0.2)\\ &=0 + 1.2 + 1.4 + 1.6 + 1.8 + 2.0\\ &=8.0 \end{align*}$$

Step3: Assess skewness

All non-zero probabilities are equal, so the distribution is symmetric (not left-skewed).

Answer:

Khaleed's claim is partially incorrect: the mean is indeed 8 interviews, but the distribution is symmetric, not skewed to the left. Overall, the full claim is not correct.