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Alex17521 [72]
3 years ago
8

According to a company's website, the top 15% of the candidates who take the entrance test will be called for an interview. You

have just been called for an interview. The reported mean and standard deviation of the test scores are 70 and 7, respectively. If test scores are normally distributed, what is the minimum score required for an interview?
Mathematics
1 answer:
galben [10]3 years ago
6 0

Answer:

The minimum score required for an interview is 77.252

Step-by-step explanation:

We solve this using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Top 15% of the candidates is a ranking that is equivalent to = 100 - 15% = 85th percentile.

The z score of 85th percentile = 1.036

Mean = 70

Standard Deviation = 7

Minimum score = raw score = ???

Hence:

1.036 = x - 70/7

Cross Multiply

1.036 × 7 = x - 70

7.252 = x - 70

x = 70 + 7.252

x = 77.252

The minimum score required for an interview is 77.252

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Interpret the interquartile range. Choose the correct answer below.
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The value 257.3 denotes the range of the middle 50% of all observations in the distribution.

Q1 = 271.8 ; Q2 = 387.9 ; Q3 = 529.1

The Interquartile range is the difference between the the upper quartile and Lower quartile ;

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The value of Q3 depict the 75th percentile of the data

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2 years ago
Problem PageQuestion Calcium levels in people are normally distributed with a mean of mg/dL and a standard deviation of mg/dL. I
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The  calcium level that is the borderline between low calcium levels and those not considered low is  c = 8.68

Step-by-step explanation:

From the question we are told that

    The mean is  \mu =  9.5\  mg/dL

     The standard deviation \sigma =  0.5 \ mg/dL

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Let X be a X random calcium level

  Now the  P(X < c) = 0.05

Here P denotes probability

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  Since the calcium level is normally distributed the z-value is  evaluated as

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The critical value for 0.05 from the standard normal distribution table is

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=>    \frac{c - \mu}{\sigma }  =  -1.645

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=>    c - 9.5 =  -0.8225

=>     c = 8.68

 

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