Math  /  Data & Statistics

QuestionUse n=6n=6 and p=0.25p=0.25 to complete parts (a) through (d) below. \begin{tabular}{l|l|} \hline 1 & 0.3560 \\ \hline 2 & 0.2966 \\ \hline 3 & 0.1318 \\ \hline 4 & 0.0330 \\ \hline 5 & 0.0044 \\ \hline 6 & 0.0002 \\ \hline \end{tabular} (Round to four decimal places as needed.) (b) Compute the mean and standard deviation of the random variable using μX=[xP(x)]\mu_{X}=\sum[x \cdot P(x)] and σx=[x2P(x)]μx2\sigma_{x}=\sqrt{\sum\left[x^{2} \cdot P(x)\right]-\mu_{x}^{2}} μX=\mu_{X}= \square (Round to two decimal places as needed.) σX=\sigma_{X}= \square (Round to two decimal places as needed.) (c) Compute the mean and standard deviation, using μX=np\mu_{X}=n p and σX=np(1p)\sigma_{X}=\sqrt{n p(1-p)}. μX=\mu_{X}= \square (Round to two decimal places as needed.) σx=\sigma_{x}= \square (Round to two decimal places as needed.)

Studdy Solution

STEP 1

1. The random variable X X represents the number of successes in a binomial experiment with n=6 n = 6 trials and probability of success p=0.25 p = 0.25 .
2. The provided probabilities correspond to the number of successes from 1 to 6.
3. The mean and standard deviation need to be calculated using both the probability distribution and the binomial formulas.

STEP 2

1. Calculate the mean using the probability distribution.
2. Calculate the standard deviation using the probability distribution.
3. Calculate the mean using the binomial formula.
4. Calculate the standard deviation using the binomial formula.

STEP 3

Calculate the mean using the probability distribution:
μX=[xP(x)]=10.3560+20.2966+30.1318+40.0330+50.0044+60.0002\mu_X = \sum[x \cdot P(x)] = 1 \cdot 0.3560 + 2 \cdot 0.2966 + 3 \cdot 0.1318 + 4 \cdot 0.0330 + 5 \cdot 0.0044 + 6 \cdot 0.0002
μX=0.3560+0.5932+0.3954+0.1320+0.0220+0.0012=1.4998\mu_X = 0.3560 + 0.5932 + 0.3954 + 0.1320 + 0.0220 + 0.0012 = 1.4998
Round to two decimal places:
μX=1.50\mu_X = 1.50

STEP 4

Calculate the standard deviation using the probability distribution:
σX=[x2P(x)]μX2\sigma_X = \sqrt{\sum\left[x^{2} \cdot P(x)\right] - \mu_X^2}
Calculate [x2P(x)]\sum[x^2 \cdot P(x)]:
=120.3560+220.2966+320.1318+420.0330+520.0044+620.0002= 1^2 \cdot 0.3560 + 2^2 \cdot 0.2966 + 3^2 \cdot 0.1318 + 4^2 \cdot 0.0330 + 5^2 \cdot 0.0044 + 6^2 \cdot 0.0002
=0.3560+1.1864+1.1862+0.5280+0.1100+0.0072=3.3738= 0.3560 + 1.1864 + 1.1862 + 0.5280 + 0.1100 + 0.0072 = 3.3738
Calculate σX\sigma_X:
σX=3.3738(1.4998)2=3.37382.2494=1.12441.0603\sigma_X = \sqrt{3.3738 - (1.4998)^2} = \sqrt{3.3738 - 2.2494} = \sqrt{1.1244} \approx 1.0603
Round to two decimal places:
σX=1.06\sigma_X = 1.06

STEP 5

Calculate the mean using the binomial formula:
μX=np=60.25=1.50\mu_X = n \cdot p = 6 \cdot 0.25 = 1.50

STEP 6

Calculate the standard deviation using the binomial formula:
σX=np(1p)=60.250.75\sigma_X = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{6 \cdot 0.25 \cdot 0.75}
σX=1.1251.0607\sigma_X = \sqrt{1.125} \approx 1.0607
Round to two decimal places:
σX=1.06\sigma_X = 1.06
The calculated values are: μX=1.50,σX=1.06\mu_X = 1.50, \quad \sigma_X = 1.06

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