Question

In a recent year, scores on a standardized test for high school students with a 3.50 to 4.00 grade point average were normally distributed, with a mean of 38.238.2 and a standard deviation of 2.22.2. A student with a 3.50 to 4.00 grade point average who took the standardized test is randomly selected. (a) Find the probability that the student's test score is less than 3737. The probability of a student scoring less than 3737 is 0.29120.2912. (Round to four decimal places as needed.) (b) Find the probability that the student's test score is between 35.135.1 and 41.341.3. The probability of a student scoring between 35.135.1 and 41.341.3 is nothing. (Round to four decimal places as needed.)

Answer #1

Part a

We have to find P(X<37)

We are given µ = 38.2, σ = 2.2

Z = (X - µ)/σ

Z = (37 - 38.2)/2.2

Z = -0.545454545

P(Z<-0.55) = 0.2912

(by using z-table or excel)

**Required probability = 0.2912**

Part b

Here, we have to find P(35.1<X<41.3)

P(35.1<X<41.3) = P(X<41.3) - P(X<35.1)

Find P(X<41.3)

Z = (X - µ)/σ

We are given µ = 38.2, σ = 2.2

Z = (41.3 - 38.2)/2.2

Z =1.409091

P(Z<1.41) = P(X<41.3) = 0.92073

(by using z-table)

Find P(X<35.1)

Z = (X - µ)/σ

Z = (35.1 - 38.2)/2.2

Z = -1.40909

P(Z<-1.41) = P(X<35.1) = 0.07927

(by using z-table)

P(35.1<X<41.3) = P(X<41.3) - P(X<35.1)

P(35.1<X<41.3) = 0.92073 - 0.07927

P(35.1<X<41.3) = 0.84146

**Required probability = 0.8415**

The probability of a student scoring between 35.135.1 and 41.341.3 is 0.8415.

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