I need help with questions 4 and 5 Use the Empirical Rule to find the following probabilities. It is suggested thatyou draw a normal curve, label the r-values for 1, 2, and 3 standard deviationsfrom the mean, and write in the probabilities between each standard deviationbefore attempting to answer the questions.The average life-span of a True Test auto battery is 39 months with a standarddeviation of 3 months. Assume a normal distribution.1. We can expect 95% of the batteries to last between33and45months.2. We can expect15% of the batteries to last less than 30months.3. We can expect 16% of the batteries to last longer than 42months.4. We can expect% of the batteries to last between 30 and39 months.5. We can expect% of the batteries to last between 42 and48 months.the question

As the distribution is symmetric, we know that if μ is the mean of the distribution, then P(X≤μ)=0.5. Empirical rule: Under this rule, 68% of the data falls within one standard deviation, 95% percent within two standard deviations, and 99.7% within three standard deviations from the mean. Here, mean and standard deviation is 39 and 3 respectively. Then, we know that P(39-(3×3)≤X≤39-(3×3))=P(30≤X≤48)=0.9973. Hence, we have P(X≤30)+P(X≥48)=1-0.9973=0.0027Also, as X is symmetric, P(X≤30)=P(X≥48)Hence, P(X≤30)=0.00272=0.00135=P(X≥48) 4.) P(30≤X≤39)=P(X≤39)-P(X≤30)=0.5-0.00135=0.49865 Hence, we can expect 49.865% of batteries to be between 30 and 39 months. 5.) P(42≤X≤48)=P(X≤48)-P(X≤42)P(X≤48)=0.5+0.99732=0.99865P(X≤42)=P(X≤36)+P(36≤X≤42)=P(X≤36)+0.68P(X≤36)=1-0.682=0.16Hence, P(X≤42)=0.16+0.68=0.84Thus,P(42≤X≤48)=0.99865-0.84=0.15865 Hence, we can expect 15.865% of batteries to be between 42 and 48 months.