In a reading test for eight-year-old children, it is found that a reading score X is normally distributed with mean 5.0 and standard deviation 2.0. (i) What proportion of children would you expect to score between 4.5 and 6.0? (ii) There are about 700 000 eight-year-olds in the country. How many would you expect to have a reading score of more than twice the mean? (iii) Why might educationalists refer to the reading score X as a ‘score out of 10’? The reading score is often reported, after scaling, as a value Y which is normally distributed, with mean 100 and standard deviation 15. Values of Y are usually given to the nearest integer. (iv) Find the probability that a randomly chosen eight-year-old gets a score, after scaling, of 103. (v) What range of Y scores would you expect to be attained by the best 20% of readers?

 In a reading test for eight-year-old children, it is found that a reading score X is normally distributed with mean 5.0 and standard deviation 2.0.

 (i) What proportion of children would you expect to score between 4.5 and 6.0?

(ii) There are about 700 000 eight-year-olds in the country. How many would you expect to have a reading score of more than twice the mean?

(iii) Why might educationalists refer to the reading score X as a ‘score out of 10’? The reading score is often reported, after scaling, as a value Y which is normally distributed, with mean 100 and standard deviation 15. Values of Y are usually given to the nearest integer.

(iv) Find the probability that a randomly chosen eight-year-old gets a score, after scaling, of 103.

 (v) What range of Y scores would you expect to be attained by the best 20% of readers?

1 Approved Answer

Manish M answered on April 21, 2021

5 Ratings,(9 Votes)

Tne formula used here is z = X - Mean/Standard Deviation Mean = 5.0 Standard Deviation = 2.0 i. To calculate the proportion of children with a score between 4.5 and 6.0, we will have to find the area under normal distribution curve to the right of z corresponding to X = 4.5 and 6.0 and find their difference. so we first calculate z value for X= 4.5 Z = 4.5-5.0/2.0 = - 0.25 From normal distribution curve the area to the right of z = - 0.25 is obtained as = 0.5987 now we calculate z value for X= 6.0 Z = 6.0-5.0/2.0 = 0.5 From normal distribution curve the area to the right of z = 0.5 is obtained as = 0.3085 Difference = 0.5987 - 0.3085 = 0.2902 so the the proportion of children with a score between 4.5 and 6.0 = 0.2902 or 29.02% ii. Number of students have a reading score of more than twice the mean: X = 2*5.0 = 10.0 we will have to find the area under normal distribution curve to the right of z corresponding to X = 10 z = X - Mean/Standard Deviation = Z = 10.0-5.0/2.0 = 2.5 From normal distribution curve the area to the right of z = 2.5 is obtained as = 0.00621 Number of students out of 700000 having a reading score of more than twice the mean = 700000* 0