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.