13.2k views
1 vote
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?

1 Answer

3 votes

This question was not written properly

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?

Answer:

0.68269

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

User Souravlahoti
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.