We gave a national verbal proficiency test that has a mean score of 500 with a standard deviation of 75. What raw score would co
1 answer:
Answer:
519
Step-by-step explanation:
Let's assume that the scores for the national verbal proficiency test follow a normal distribution.
Mean score (μ) = 500
Standard deviation (σ) = 75
According to a Z-score table, the score for the 60th percentile is roughly z = 0.253
For any score "x", the z-score is given by:
For z = 2.53:
The raw score to the 60th percentile is 519.
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Answer:
The probability of Steve agreeing with the company’s claim is 0.50502.
Step-by-step explanation:
Let <em>X</em> denote the number of green candies.
The probability of green candies is, <em>p</em> = 0.20.
Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.
So, <em>n</em> = 30 candies are randomly selected.
All the candies are independent of each other.
The random variable <em>X</em> follows a binomial distribution with parameter <em>n</em> = 30 and <em>p</em> = 0.20.
It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.
Compute the probability of 5, 6 and 7 green candies as follows:
Then the probability of Steve agreeing with the company’s claim is:
P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)
= 0.17228 + 0.17946 + 0.15328
= 0.50502
Thus, the probability of Steve agreeing with the company’s claim is 0.50502.
Answer:
From the frequency table, let's calculate the row total.
Row total for phone call = 19 + 9 = 28
Row total for no phone call = 8 +6 = 14
To calculate their respective row relative frequencies, let's use:
Row relative freq =
Now, the two-way frequency table will be computed as:
For phone call:
Desirable behavior = ≈0.69
Undesirable behaviour = ≈0.32
No phone call:
Desirable behaviour = ≈ 0.57
Undesirable behaviour = ≈ 0.43
The complete two-way table is attached.
