The mean of a normal probability distribution is 60; the standard deviation is 5. a. About what percent of the observations lie between 55 and 65? b. About what percent of the observations lie between 50 and 70? c. About what percent of the observations lie between 45 and 75?

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Zachary Pickett · Accepted Answer

Step 1    Mean of the normal distribution is μ=60 and the standard deviation is σ=5    a. Observations lie between 55 and 65    Suppose that area under the normal curve with these observations is within z standard deviation of the mean.    ⇒μ−zσ=55 and μ+zσ=65    ⇒60−5z=55 and 60+5z=65    ⇒5z=60−55 and 5z=65−60    ⇒5z=5 and 5z=5    ⇒z=1 and z=1    As per the empirical rule, about 68% of the area under the normal curve is within one standard  deviation of the mean.

Harr1957 · Accepted Answer

b. Observations lie between 50 and 70    Suppose that area under the normal curve with these observations is within z standard deviation of the mean.    ⇒μ−zσ=50 and μ+zσ=70    ⇒60−5z=50 and 60+5z=70    ⇒5z=60−50 and 5z=70−60    ⇒5z=10 and 5z=10    ⇒z=2 and z=2    As per the empirical rule, about 95% of the area under the normal curve is within two standard  deviation of the mean.

user_27qwe · Accepted Answer

Step 3 c. Observations lie between 45 and 75 Suppose that area under the normal curve with these observations is within z standard deviation of the mean. ⇒μ−zσ=45 and μ+zσ=75 ⇒60−5z=45 and 60+5z=75 ⇒5z=60−45 and 5z=75−60 ⇒5z=15 and 5z=15 ⇒z=3 and z=3 As per the empirical rule, Practically all of the area under the normal curve is within three standard deviation of the mean.

Don Sumner · Accepted Answer

a. We want to calculate P(55≤X≤65), where X is a random variable following a normal distribution with mean μ and standard deviation σ.Using the standard normal distribution, we can standardize the values 55 and 65 by subtracting the mean and dividing by the standard deviation:z1=55−605=−1z2=65−605=1Next, we can find the corresponding cumulative probabilities for these standardized values. The cumulative distribution function (CDF) of the standard normal distribution is denoted by Φ(z). Therefore, we need to calculate P(−1≤Z≤1), where Z is a standard normal random variable.Using the properties of the standard normal distribution, we can find these probabilities:P(−1≤Z≤1)=Φ(1)−Φ(−1)You can use a standard normal distribution table or a calculator to find the values of Φ(1) and Φ(−1). Subtracting these values will give you the required probability.b. To find the percentage of observations between 50 and 70, we follow a similar approach as in part a.z1=50−605=−2z2=70−605=2P(−2≤Z≤2)=Φ(2)−Φ(−2)c. To find the percentage of observations between 45 and 75:z1=45−605=−3z2=75−605=3P(−3≤Z≤3)=Φ(3)−Φ(−3)

nick1337 · Accepted Answer

Step 1:a. To find the percentage of observations lying between 55 and 65 in a normal distribution with a mean of 60 and a standard deviation of 5, we can use the cumulative distribution function (CDF) of the normal distribution. The CDF gives the probability of a value being less than or equal to a given point.Let X be a random variable following a normal distribution with mean μ=60 and standard deviation σ=5. We want to find P(55≤X≤65).Using the standardization formula, we can transform the values of 55 and 65 into their corresponding z-scores:z1=55−μσ=55−605=−1z2=65−μσ=65−605=1Next, we can use the standard normal distribution table or a calculator to find the probabilities associated with these z-scores. From the table, we find that P(Z≤−1)≈0.1587 and P(Z≤1)≈0.8413.The probability of the observations lying between 55 and 65 is given by:P(55≤X≤65)=P(−1≤Z≤1)=P(Z≤1)−P(Z≤−1)=0.8413−0.1587=0.6826Therefore, approximately 68.26% of the observations lie between 55 and 65.Step 2:b. To find the percentage of observations lying between 50 and 70, we can follow a similar approach as in part (a).Let X be a random variable following a normal distribution with mean μ=60 and standard deviation σ=5. We want to find P(50≤X≤70).Using the standardization formula, we calculate the corresponding z-scores:z1=50−μσ=50−605=−2z2=70−μσ=70−605=2Using the standard normal distribution table or a calculator, we find P(Z≤−2)≈0.0228 and P(Z≤2)≈0.9772.The probability of the observations lying between 50 and 70 is:P(50≤X≤70)=P(−2≤Z≤2)=P(Z≤2)−P(Z≤−2)=0.9772−0.0228=0.9544Therefore, approximately 95.44% of the observations lie between 50 and 70.Step 3:c. To find the percentage of observations lying between 45 and 75, we follow the same steps as in the previous parts.Let X be a random variable following a normal distribution with mean μ=60 and standard deviation σ=5. We want to find P(45≤X≤75).Using the standardization formula, we calculate the z-scores:z1=45−μσ=45−605=−3z2=75−μσ=75−605=3From the standard normal distribution table or a calculator, we find P(Z≤−3)≈0.0013 and P(Z≤3)≈0.9987.The probability of the observations lying between 45 and 75 is:P(45≤X≤75)=P(−3≤Z≤3)=P(Z≤3)−P(Z≤−3)=0.9987−0.0013=0.9974Therefore, approximately 99.74% of the observations lie between 45 and 75.

The mean of a normal probability distribution is 60; the standard deviation is 5

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