Question

Enter #22 Points possible: 1 . Total attempts: 3 7.3 Central Limit Theorem - Finding X Values The lengths of pregnancies in a small rural village are normally distributed with a mean of 260.4 days and a standard deviation of 17.2 days. In what range would you expect to find the middle 98% of most pregnancies? Between and Enter your answers as numbers. Your answers should be accurate to 1 decimal places.

Answer 1

X : Length of pregenancy

X follows normal ditribution with mean 260.4 days and standard deviation of 17.2 days.

Let X1 be lower value of the range of length of pregenancy X2 be the upper value of the range for length of pregancy. such that length of pregenancy between X1 and X2 are 98%

From the above picture, X1 : Length of the pregnancy such that 1%of the pregenancies where length of pregenacy is < X1 i.e P(X<X1) = 0.01

Similary X2: Length of the pregnancy such that 1%of the pregenancies where length of pregenacy is > X2 i.e P(X>X2) = 0.01

P(X<X1) = 0.01

Let Z1 be the Z-score for X1

Z1 = (X1 - mean)/ Standard deviation =(X1-260.4)/17.2

P(X<X1) = 0.01 ie. P(Z<Z1) = 0.01

From standard normal tables,

Value of Z1 such that P(Z<Z1): 0.01 = -2.33

Z1=-2.33

Z1 = (X1-260.4)/17.2

-2.33 x 17.2 = X1 - 260.4;

X1 = 260.4 - 2.33 x 17.2 = 260.4 - 40.076 = 220.324

X1 = 220.324

Similary X2: Length of the pregnancy such that 1%of the pregenancies where length of pregenacy is > X2 i.e P(X>X2) = 0.01

P(X>X2) = 0.01

P(X>X2) = 1-P(X<X2) = 0.01 ; P(X<X2) = 1-0.01 = 0.99

Z2 : Zscore for X2 ; Z2 = (X2 - 260.4)/17.2

P(X<X2) = 0.99 ie. P(Z<Z2) = 0.99

From standard normal tables,

Value of Z2 such that P(Z<Z2): 0.99 = 2.33

Z2 = 2.33

Z2 = (X2-260.4)/17.2

2.33 x 17.2 = X2 - 260.4;

X2 = 260.4 + 2.33 x 17.2 = 260.4 + 40.076 = 300.476

X2 = 300.476

X1 = 220.324 ; X2 = 300.476

Range of Middle 98% of most pregenancies between 220.324 dayss and 300.476 days.