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The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen's School of Business in Kingston, Ontario requires a minimum LSAT score of 157 for admission. What is the z-score for a score of 157? Round to three decimal places (Example: 0.398).

Respuesta :

Answer:the z score is 0.618

Step-by-step explanation:

Since the scores on the LSAT are approximately normal, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores on the LSAT.

µ = mean score

σ = standard deviation

From the information given,

µ = 150.7

σ = 10.2

We want to find the z-score for a score of 157

For x = 157,

z = (157 - 150.7)/10.2 = 6.3/10.2

z = 0.618

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