Question

132 and standard deviation o35. [ You may find it useful to reference the z table.) Let Xbe normally distributed with mean a. Find PXs 100). (Round " value to 2 decimal places and final answer to 4 decimal places.) P(Xs 100) value to 2 decimal places and final answer to 4 decimal places.) b. Find P95 s Xs 110). (Round " P(95 sXs 110) value and final answer to 3 decimal places.) c. Find x such that PXs x) 0.350. (Round " value and final answer to 3 decimal places.) 0.810. (Round" d. Find x such that PX> Novt

Answer 1

Solution:-

Given that,

Mean = = 132

Standard deviation = =35

(A).

P(x 100)

= P [(x - ) / (100 - 132) / 35]

= P (z -0.91)

= 0.1814

P(x 100) = 0.1814

(B).

= P [(95 - 132 / 35) (x - ) / (110 - 132 / 35) ]

= P (-1.06 z -0.62)

= P (z -0.63) - P(z -1.06)

= 0.2643 - 0.1446

= 0.1197

P (95 X 110) = 0.1197

(C).

Using standard normal table,

P (Z < z) = 0.350

P (Z < -0.1) = 0.350

z = -0.39

Using z-score formula,

x = z * +

x = -0.39 * 35 + 132 = 118.35

(D).

Using standard normal table ,

P(Z > z) = 0.810

1 – P (Z < z) = 0.810

P (Z < z) = 1 - 0.810

P (Z < -0.87) = 0.19

z = -0.87

Using z-score formula,

x = z * +

x = -0.87 * 35 + 132 = 101.55