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A group of students at Brunel university took a test in marketing and they needed to score 50 to pass. The results showed that the final grades have a mean of 62 and a standard deviation of 8. Suppose we approximate the distribution of these grades by normal distribution. Based on this information, you are required to perform the following; a) what percetage(to 3 decimals) of the students scored more than 78? B) what percentage of the students to 3 decimal places scored under 50? And c) what percentage to 3 decimal places of the students scored between 60 and 80?

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mtngkhoi

Answer:

My answer: 15.87%

Step-by-step explanation:

The Final grades follows normal distribution.

It means =70

The Given standard deviation =10

The normal random variable of a standard normal distribution is called a standard score or a z-score. In Every normal condition, random variable X will be transformed into a z score via the following equation:

z=(X)

But when X is a normal random variable, μ is the mean of X.

For That, it is X=60

Z=(60−70)/10=−1

P(X<60)=P(Z<−1)

=0.1587

Hence percent of students failed in test is 15.87%

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