wild deer population weight has standard deviation of 50 lb. What is the probability of existence of a random sample (size 100) weight within + 5 lb. of population mean weight?

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Complete Question

Wild deer population weight has standard deviation of 50 lb. What is the probability of existence of a random sample (size 100) weight within ± 5 lb. of population mean weight?

0.68269

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ/√n where

x is the raw score

μ is the population mean

σ is the population standard deviation.

From the question:

Standard deviation = 50 lb.

Sample size = 100

The weight is within ± 5Ib

For +5 lb

Z =+ 5/50/ √100

Z = +5/50/10

Z = +5/5

Z = 1

Using the Z table to find the probability of the Z score.

P(Z = 1) = 0.84134

For - 5 lb

Z = - 5/50/ √100

Z = - 5/50/10

Z = - 5/5

Z = -1

Using the Z table to find the probability of the Z score.

P(Z = -1 ) = 0.15866

Hence,

P (-Z<x<Z)

P(Z = +1 ) - P(Z = -1)

0.84134 - 0.15866

= 0.68269

The probability of existence of a random sample (size 100) weight within ± 5 lb. of population mean weight is 0.68269

47 - 90 = 43

43 - 180 = x

x = 137