Complete Question
Given a population where the probability of success is p= 0.40 calculate the probabilities below if a sample of 300 is taken.
A. Calculate the probability the proportion of successes in the sample will be less than 0.42 (round 4 decimals)
B. What is the probability that the proportion of successes in the sample will be greater than 0.44 (round 4 decimals)
Answer:
A
[tex]P(X < 0.42) = 0.76028[/tex]
B
[tex]P(X > 0.44) = 0.078622[/tex]
Step-by-step explanation:
From the question we are told that
The probability of success is p = 0.40
The sample size is n = 300
Generally given that the sample size is large enough n > 30 then the mean for this sampling distribution is
[tex]\mu_{x} = p = 0.40[/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{ \frac{p (1 - p )}{n} }[/tex]
=> [tex]\sigma = \sqrt{ \frac{0.40 (1 - 0.40 )}{ 300} }[/tex]
=> [tex]\sigma = 0.02828[/tex]
Considering question A
Generally the probability the proportion of successes in the sample will be less than 0.42 is mathematically represented as
[tex]P(X < 0.42) = P(\frac{X - \mu }{\sigma } < \frac{0.42 - 0.40 }{ 0.02828} )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(X < 0.42) = P(Z < 0.7072 )[/tex]
From the z table
The area under the normal curve to the left corresponding to 0.7072 is
[tex]P(Z < 0.7072 ) = 0.76028[/tex]
=> [tex]P(X < 0.42) = 0.76028[/tex]
Considering question B
Generally the probability the proportion of successes in the sample will be less than 0.44 is mathematically represented as
[tex]P(X > 0.44) = P(\frac{X - \mu }{\sigma } > \frac{0.44 - 0.40 }{ 0.02828} )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(X > 0.44) = P(Z > 1.4144 )[/tex]
From the z table
The area under the normal curve to the left corresponding to 1.4144 is
[tex]P(Z > 1.4144 ) = 0.078622[/tex]
=> [tex]P(X > 0.44) = 0.078622[/tex]
