An account earns annual simple interest. Interest earned: $84 principal:$600 interest rate:7% find the time ?
1 answer:
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<u>Answer:</u>
The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.
<u>Solution:</u>
Given, Scores on a test are normally distributed with a mean of 81.2
And a standard deviation of 3.6.
We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?
For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6
Now probability of getting more than 88.4 = 1 - area of z – score of 88.4
So, probability of getting more than 88.4 = 1 – area of z- score(2)
= 1 – 0.9772 [using z table values]
= 0.0228.
Now probability of getting more than 77.6 = 1 - area of z – score of 77.6
So, probability of getting more than 77.6 = 1 – area of z- score(-1)
= 1 – 0.1587 [Using z table values]
= 0.8413
Now, probability of getting in between 77.6 and 88.4 = 0.8413 – 0.0228 = 0.8185
Hence, the probability of a randomly selected student getting in between 77.6 and 88.4 is 0.8185.